Solved questions from Physics Grade 12 previously asked in exams

 

Physics

Grade 12 ,

2071 Q.NO. 1 (c) REGULAR.

Sound travels faster in metal than in air, why?  

 

[Marks : 2 ] 

Ans. :

The sound is longitudinal wave. In general the velocity of longitudinal wave is given by v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$, where γ is the modulus of elasticity and ρ is the density of the medium. For the solids, the ratio (γ/ρ) is much greater than that for gases. Hence, the velocity of sound in solids is greater than in gases, although the densities of solids are greater than that of gases.

2071 Q.NO. 11 REGULAR.

A source of sound of frequency 512Hz emits waves of wavelength 64.5cm in air at 200C. what would be the velocity of sound at 00C? 

 

 

[Marks : 4 ] 

Ans. :

.Solution.

Frequency of sound (f)=512H

Wavelength of sound at 200C(λ20)=64.5cm=6.45 x 10-1m

Temperature at200C (T20)=200C=20+273=293K

Temperature 00C(T0)=00C=0+273=273K

Velocity of sound in at 200C(V20)=?

Velocity of sound in air at 00C(V0)=?

We know, v=λf

Or, v20=λ20f=6.45 x 10-1x 512=330.24 m/sec

Since the velocity is directly proportional to the square root of temperature the we have

V0V20=T0T20−−−√$\frac{V_0}{V_{20}}=\sqrt{\frac{T_0}{T_{20}}}$

Or,V0=V20$V_0=V_{20}$T0T20−−−√$\sqrt{\frac{T_0}{T_{20}}}$=330.24 x 273293−−−√$\sqrt{\frac{273}{293}}$=318.8 m/sec.

2071 Q.NO. 7 (a) SUPPLEMENTARY.

Describe the Newton formula for velocity of sound in air with Laplace’s correction.

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Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2070 Q.NO. REGULAR.

What  is the principle of the superposition as applied to wave motion? Discuss the result of superposing two wave of the equal amplitude and same frequency travelling in opposite direction.

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Ans. :

When two wave trains of same frequency and amplitude travel with the same velocity along the same straight line in opposite directions, they superimpose and produce a new type of wave called stationary wave or standing wave.

The name stationary for such type of waves is justified because there is no flow of energy along the wave. Let the incident wave propagating along Y- axis be

25.  

and the wave reflected from the boundary traveling along negative X-axis is

25.  

when two wave superimposed then the stationery wave is produce then the resultant wave is given by y=y1 +y2=a sin(ωt+kx)+asin(ωt-kx)

or, y=sinωt Cos (-kx)=2a coskx. Sinωt

or, y=2a cos2πλ$\frac{2\pi }{\lambda }$x sinωt……..3

eq.3 represent the equation of stationary wave.

Here 2acoskx is amplitude and sinωt give the nature of the amplitude of the oscilliaration.

Special cases

If 2a coskx=0 then coskx=0, therefore x=(n+12)$\left(\mathrm{n}+\frac{1}{2}\right)$∏

Or, x=(n+12)λ2$\frac{\left(\mathrm{n}+\frac{1}{2}\right)\lambda }{2}$  where n is the interger. The point x is always at rest called node.

When n = 0

X1 = λ /4

when  n = 1

x2 = 34$\frac{3}{4}$ λ 

The difference between two different nodes x2  –   x1 =  λ/2

 

If 2a coskx =±1$\pm 1$ then, kx=n∏

Or, x=nλ/2

Here at the point x maximum displacement occur called antinodes.

2070 Q.NO. REGULAR.

What is stationary wave? Prove that the distance between any two consecutive nodes in a stationary wave is λ/2. 70 7a

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Ans. :

When two wave trains of same frequency and amplitude travel with the same velocity along the same straight line in opposite directions, they superimpose and produce a new type of wave called stationary wave or standing wave.

The name stationary for such type of waves is justified because there is no flow of energy along the wave. Let the incident wave propagating along Y- axis be

25.  

and the wave reflected from the boundary traveling along negative X-axis is

25.  

when two wave superimposed then the stationery wave is produce then the resultant wave is given by y=y1 +y2=a sin(ωt+kx)+asin(ωt-kx)

or, y=sinωt Cos (-kx)=2a coskx. Sinωt

or, y=2a cos2πλ$\frac{2\pi }{\lambda }$x sinωt……..3

eq.3 represent the equation of stationary wave.

Here 2acoskx is amplitude and sinωt give the nature of the amplitude of the oscilliaration.

Special cases

If 2a coskx=0 then coskx=0, therefore x=(n+12)$\left(\mathrm{n}+\frac{1}{2}\right)$∏

Or, x=(n+12)λ2$\frac{\left(\mathrm{n}+\frac{1}{2}\right)\lambda }{2}$  where n is the interger. The point x is always at rest called node.

When n = 0

X1 = λ /4

when  n = 1

x2 = 34$\frac{3}{4}$ λ 

The difference between two different nodes x2  –   x1 =  λ/2

 

If 2a coskx =±1$\pm 1$ then, kx=n∏

Or, x=nλ/2

Here at the point x maximum displacement occur called antinodes.

2070 Q.NO. REGULAR.

How is a progressive wave different from a stationary wave? Derive a progressive wave equation.

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Ans. :

Progressive Wave: 

1. The wave profile is seen to move indicating that energy is being transported in the process.

 2. All points or particle will oscillated with the same amplitude or constant amplitude.

3. Points within a wavelength are out of phase. Two successive points that are in phase are exactly one wavelength apart.

Stationary Wave :

1. The wave profile is stationary showing that there is no net transfer of energy from one end to another.

 2.  the Points vibrate with different amplitudes  or not constant i.e. ranging from zero  to a maximum amplitude.

3. The Points within one loop  are in phase. However, they are in anti-phase with the vibrations of the points in the adjacent loop.

In that way stationary wave and progressive wave are different.

The simplest type of wave is the one in which the particles of the medium are set into simple harmonic vibrations as the wave passes through it. The wave is then called a simple harmonic wave.

Figure: 2

 

Consider a particle O in the medium. The displacement at any instant of time is given by

y=A sinωt…………..1

Where A is the amplitude, ω is the angular frequency of the wave. Consider a particle P at a distance x from the particle O on its right. Let the wave travel with a velocity v from left to right. Since it takes some time for the disturbance to reach P, its displacement can be written as

y=A sin(ωt -ɸ)…………….2

Where ɸ is the phase difference between the particles O and P.

                Hence a path difference of x corresponds to a phase difference of 2πx/λ=ɸ

Substituting the value of the ɸ in equation 2.

We get,

y=A sin(ωt -2πx/λ )…………….3

since we have ω=2π/T=2πf

 

y=Asin{2πtT−2πxλ}$\mathrm{y}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}\left\{\frac{2\pi \mathrm{t}}{\mathrm{T}}-\frac{2\pi \mathrm{x}}{\lambda }\right\}$

y=Asin2πλ{vt−x}$\mathrm{y}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}\frac{2\pi }{\lambda }\left\{\mathrm{v}\mathrm{t}-\mathrm{x}\right\}$ ………4

Similarly, for a particle at a distance x to the left of 0, the equation for the displacement is   given by

If the wave travel from right to left in negative x-axis then the equation for the displacement is     given by

y=Asin2πλ{vt+x}$\mathrm{y}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}\frac{2\pi }{\lambda }\left\{\mathrm{v}\mathrm{t}+\mathrm{x}\right\}$ ………5

               This is the expression of the progressive wave equation.  ALL BOX ARE ɸ

2070 Q.NO. 3 (a) REGULAR.

Although te density of the solid is hight, the velocity of sound is grater in solid. Explain.

 

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Ans. :

The sound is longitudinal wave. In general the velocity of longitudinal wave is given by v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$, where γ is the modulus of elasticity and ρ is the density of the medium. For the solids, the ratio (γ/ρ) is much greater than that for gases. Hence, the velocity of sound in solids is greater than in gases, although the densities of solids are greater than that of gases.

2070 Q.NO. 7 (a) SUPPLEMENTARY.

Deduce Newton’s formula for the velocity of sound in gas. Why was the correction needed? What made by Laplace? 

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Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2069 Q.NO. REGULAR.

Why the echo cannot be heard in a small room?

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Ans. :

The phenomenon of reflection of sound from a surface is called echo. For example, when you shout from a valley, you hear an echo.

Any sound heard by human ears persists for about 0.1 second. So, to hear distinct echo the minimum distance between the source of sound and the reflector object must be 17 m. but the size of the room is very smaller than that of 17m so we cannot hear the echo at the small room.

2069 Q.NO. REGULAR.

A radio station broadcast at 800KHz. What is the wavelength of the wave?

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Ans. :

Ans. a/C to question we have given

Velocity of the wave( C)= 3 * 108m/s

Frequency of the wave(f)=800KHz=8 * 105Hz

Wavelength of the wave emitted from radio station (λ)=?

Since we have relation C=λf

Or, λ=C/f=3∗1088∗105$\frac{3\ast {10}^8}{8\ast {10}^5}$=3*103/8=375m

2069 Q.NO. 2 REGULAR.

Discuss the effect of pressure, temperature, and density of elastic medium on the velocity of sound. 

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Ans. :

The factor which is affecting the velocity of the light is given below:

a.Pressure

b.Temperature

c.Density

d.moisture or humidity

e.Frequency, wavelength and amplitude.

a.Pressure: there is no effect of the pressure in velocity of sound because the ratio of P/ρ is constant i.e. when the pressure increase then density also increases at constant temperature. On the other hand Since we have relation v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$ in this relation γ is constant  and P/ρ is also constant then from the above relation we have v=a constant. At the given temperature.

b.Temperature:  since we have a relation v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$  we also have ρ=M/V  where M be the mass of the gas V be the volume. Then v=γPVM−−−−√$\sqrt{\frac{\gamma PV}{M}}$ ………..1

Now from ideal gas equation for one mole gas we have relation PV=RT…..2

From 1 and 2 we get

v= since γ, R and ρ is constant then we have v=where C=

now from equation 3 we have relation that velocity of the sound is proportional to the square root of the absolute temperature at a given pressure.

c.Density : since we have the relation v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$  let us consider  two gases at the  same temperature then we have two relation i.e.

v1=γPρ1−−−√………….1$v_1=\sqrt{\frac{\gamma P}{{\rho }_1}}\dots \dots \dots \dots .1$ for one gas and

v2=γPρ2−−−√………….2$v_2=\sqrt{\frac{\gamma P}{{\rho }_2}}\dots \dots \dots \dots .2$ for another gas.

equation 1 and 2 we have

v1v2=ρ2ρ1−−√$\frac{v_1}{v_2}=\sqrt{\frac{{\rho }_2}{{\rho }_1}}$  in general we have v α 1ρ√$\frac{1}{\sqrt{\rho }}$

This is the relation of the velocity of the sound and the density.

This relation show that the gas which has lighter density has higher velocity and those have higher density have lower velocity

d.Humidity or moisture:  since the density of moisture gas is less than the density of dry gas and we have relation v α 1ρ√$\frac{1}{\sqrt{\rho }}$ due to this relation the velocity of moisture is greater than that of dry air.

e.Frequency or wavelength:  as we have relation v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$ in this relation frequency and wavelength is not affect the formula of velocity of sound. So that the sound of any frequency or wavelength travels through a given material with the same velocity.

2069 Q.NO. 2 (b) REGULAR.

What is Newton’s formula for the velocity of sound? What correction was made by Laplace? 

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Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2069 Q.NO. 3 (b) REGULAR.

Is it possible that the velocity of sound is grater in solid than that in gases at STP? Justify your answer. 

 

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Ans. :

The sound is longitudinal wave. In general the velocity of longitudinal wave is given by v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$, where γ is the modulus of elasticity and ρ is the density of the medium. For the solids, the ratio (γ/ρ) is much greater than that for gases. Hence, the velocity of sound in solids is greater than in gases, although the densities of solids are greater than that of gases.

2069 Q.NO. 5 (a) REGULAR.

Write an expression for the speed of sound in an ideal gas. Discuss the effect of change in the temperature on the speed of sound. 

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2069 Q.NO. 11 REGULAR.

If a detonator is exploded on a railway line, an observer standing on the rail 2km away hears two reports. What is the time interval between these two reports.[ young modulus for steel = 2x1011Nm-2, density of the steel=8x103kgm-3 , density of air =1.4  kgm-3, ratio of molar heat capacities of air=1.4, atmospheric pressure = 105Nm-2] 

 

 

[Marks : 4 ] Hide Ans

Ans. :

 solution

The separation between the rail line and observer(d)=2Km=2000m

the young’s modulus of steel(Ysteel )= 2 x 1011N/m2

Density of the steel(ρsteel )= 8 x103kg/m-3

Density of air (ρair)=1.4 kgm-3

Ratio of the molar heat capacities of air (γ)=1.4

Atmospheric pressure (P)= 105N/m-2

Now the time intravel between the two sound one heard by air medium and another heard by putting the ear on the rail line (∆t)=?

a/c to the question , ∆t= tair   - tsteel

Since the distance is same then, ………1

also we have

Vair= γPρair−−−√$\sqrt{\frac{\gamma P}{{\rho }_{air}}}$  and Vsteel=Ysteelρsteel−−−−√$\sqrt{\frac{Y_{steel}}{{\rho }_{steel}}}$

On putting these value in equation 1 we get

∆t =s{ρairγP−−−√−Ysteelρsteel−−−−√}$s\left\{\sqrt{\frac{{\rho }_{air}}{\gamma P}}-\sqrt{\frac{Y_{steel}}{{\rho }_{steel}}}\right\}$={1.41.4×105−−−−−−√−8×1032×1011−−−−−√}×2000$\left\{\sqrt{\frac{1.4}{1.4\times {10}^5}}-\sqrt{\frac{8\times {10}^3}{2\times {10}^{11}}}\right\}\times 2000$=5.92sec.

2069 Q.NO. 7 (a) SUPPLEMENTARY.

Describe Laplace correction to find the velocity of sound and discuss how the velocity of sound is affect by the difference physical parameters. 

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2069 Q.NO. 5 (a) OLD.

What is Newton’s formula for the velocity of the sound? What correction was made by Laplace?

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2068 Q.NO. REGULAR.

Can two person on the moon hear the sound of each other? Explain. 

[Marks : 4 ] Hide Ans

Ans. :

As sound required medium to travel but on the moon there is no atmosphere or it is like to be vacuum so cracking of sound is not heard behind on the moon surface.

2068 Q.NO. REGULAR.

Distinguish between light waves and sound waves. 

[Marks : 4 ] Hide Ans

Ans. :

Distinguish between light waves and sound waves.

Sound wave.	Light wave.
1. Require a medium.	1. Do not require a medium
2. Travel with a speed of 332 m/s at 0 C.	2. Travel with a speed of 3 * 108 m/s.
3. It is longitudinal waves consisting of compressions and rarefactions.	3. It is electromagnetic waves consisting of varying electric and magnetic fields.
4. It is of different frequencies, giving notes of different pitches.	4. it is of different frequencies, resulting in different colors.
5. They do not travel far as their energy is dissipated easily.	5. they can travel through a much greater distance.

2068 Q.NO. REGULAR.

What  is the principle of the superposition as applied to wave motion? Discuss the result of superposing two wave of the equal amplitude and same frequency travelling in opposite direction.

[Marks : 10 ] Hide Ans

Ans. :

When two wave trains of same frequency and amplitude travel with the same velocity along the same straight line in opposite directions, they superimpose and produce a new type of wave called stationary wave or standing wave.

The name stationary for such type of waves is justified because there is no flow of energy along the wave. Let the incident wave propagating along Y- axis be

25.  

and the wave reflected from the boundary traveling along negative X-axis is

25.  

when two wave superimposed then the stationery wave is produce then the resultant wave is given by y=y1 +y2=a sin(ωt+kx)+asin(ωt-kx)

or, y=sinωt Cos (-kx)=2a coskx. Sinωt

or, y=2a cos2πλ$\frac{2\pi }{\lambda }$x sinωt……..3

eq.3 represent the equation of stationary wave.

Here 2acoskx is amplitude and sinωt give the nature of the amplitude of the oscilliaration.

Special cases

If 2a coskx=0 then coskx=0, therefore x=(n+12)$\left(\mathrm{n}+\frac{1}{2}\right)$∏

Or, x=(n+12)λ2$\frac{\left(\mathrm{n}+\frac{1}{2}\right)\lambda }{2}$  where n is the interger. The point x is always at rest called node.

When n = 0

X1 = λ /4

when  n = 1

x2 = 34$\frac{3}{4}$ λ 

The difference between two different nodes x2  –   x1 =  λ/2

 

If 2a coskx =±1$\pm 1$ then, kx=n∏

Or, x=nλ/2

Here at the point x maximum displacement occur called antinodes.

2068 Q.NO. 7 (a) REGULAR.

Discuss the Newton formula for the velocity of the sound in air with Laplace correction.

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2067 Q.NO. 2 OLD.

 Discuss the Laplace’s correction for the velocity of the sound in air. 

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2067 Q.NO. 5 (b) OLD.

At what temperature the velocities of sound in air increase by 40% to that at 270C? 

 

 

[Marks : 4 ] Hide Ans

Ans. :

solution

Temperature (T27)=270=27+273=300K

Velocity of the sound at 270(V27)= v

Temperature at which velocity of air is increase by 40%(T)=?

Velocity that T temperature (VT)=v + v of 40%=1.4v

Since we have relation

Or, T= VTV27T27−−−√$\frac{V_T}{V_{27}}\sqrt{T_{27}}$= 1.4vv300−−−√$\frac{1.4v}{v}\sqrt{300}$ = 588K.

2066 Q.NO. REGULAR.

Define progressive wave. Derive progressive wave equation.

[Marks : 10 ] Hide Ans

Ans. :

Progressive Wave: 

1. The wave profile is seen to move indicating that energy is being transported in the process.

 2. All points or particle will oscillated with the same amplitude or constant amplitude.

3. Points within a wavelength are out of phase. Two successive points that are in phase are exactly one wavelength apart.

Stationary Wave :

1. The wave profile is stationary showing that there is no net transfer of energy from one end to another.

 2.  the Points vibrate with different amplitudes  or not constant i.e. ranging from zero  to a maximum amplitude.

3. The Points within one loop  are in phase. However, they are in anti-phase with the vibrations of the points in the adjacent loop.

In that way stationary wave and progressive wave are different.

The simplest type of wave is the one in which the particles of the medium are set into simple harmonic vibrations as the wave passes through it. The wave is then called a simple harmonic wave.

Figure: 2

 

Consider a particle O in the medium. The displacement at any instant of time is given by

y=A sinωt…………..1

Where A is the amplitude, ω is the angular frequency of the wave. Consider a particle P at a distance x from the particle O on its right. Let the wave travel with a velocity v from left to right. Since it takes some time for the disturbance to reach P, its displacement can be written as

y=A sin(ωt -ɸ)…………….2

Where ɸ is the phase difference between the particles O and P.

                Hence a path difference of x corresponds to a phase difference of 2πx/λ=ɸ

Substituting the value of the ɸ in equation 2.

We get,

y=A sin(ωt -2πx/λ )…………….3

since we have ω=2π/T=2πf

 

y=Asin{2πtT−2πxλ}$\mathrm{y}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}\left\{\frac{2\pi \mathrm{t}}{\mathrm{T}}-\frac{2\pi \mathrm{x}}{\lambda }\right\}$

y=Asin2πλ{vt−x}$\mathrm{y}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}\frac{2\pi }{\lambda }\left\{\mathrm{v}\mathrm{t}-\mathrm{x}\right\}$ ………4

Similarly, for a particle at a distance x to the left of 0, the equation for the displacement is   given by

If the wave travel from right to left in negative x-axis then the equation for the displacement is     given by

y=Asin2πλ{vt+x}$\mathrm{y}=\mathrm{A}\mathrm{s}\mathrm{i}\mathrm{n}\frac{2\pi }{\lambda }\left\{\mathrm{v}\mathrm{t}+\mathrm{x}\right\}$ ………5

               This is the expression of the progressive wave equation.  ALL BOX ARE ɸ

2066 Q.NO. 1 (c) REGULAR.

Velocity of sound increase on a cloudy day. Why? 

 

 

[Marks : 2 ] Hide Ans

Ans. :

Velocity of sound is more in damp air or moisture air. We know that the density of damp air is less than air due to the presence of humidity. Also we have the relation (v α )i.e. the velocity of sound in air is inversely proportional to the square root of the density of the medium. So the velocity of sound in damp air is more than that of in dry air.

2065 Q.NO. 5 (a) REGULAR.

Explain the significance of the Laplace correction of a Newton’s formula for the velocity of the sound and derive the correct formula. 

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2065 Q.NO. 5 (a') REGULAR.

Derive an expression for the velocity of sound in a medium by dimensional method. Discuss the effect of change in pressure and temperature on the velocity of sound in air. 

 

[Marks : 4 ] Hide Ans

Ans. :

Ans. let us consider v be the velocity of sound and E be the elasticity of the medium in which the sound travel and ρ be the density of the medium.

Then v

Or, v=k……….1

Or, v=kEaρb$E^a{\rho }^b$……….1

Or, [LT−1]=k[MLT−2/L2]a[ML−3]b$\left[L{T}^{-1}\right]=k{\left[ML{T}^{-2}/L^2\right]}^a{\left[M{L}^{-3}\right]}^b$

Or, [LT−1]=k[MT−2L−1]a[ML−3]b$\left[L{T}^{-1}\right]=k{\left[M{T}^{-2}{L}^{-1}\right]}^a{\left[M{L}^{-3}\right]}^b$

Or, [LT−1]=k[Ma+bL−a−3bT−2a]$\left[L{T}^{-1}\right]=k\left[M^{a+b}{L}^{-a-3b}{T}^{-2a}\right]$

Equating the power of M, L, and T we have

a+b=0, -a-3b=1 and -1=-2a on solving this expression we get

a=1/2, b=-1/2 putting this value on equation 1 we get

v=E1/2.ρ-1/2=.

This is the expression of the velocity of sound in a medium by dimensional method.

Pressure: there is no effect of the pressure in velocity of sound because the ratio of P/ρ is constant i.e. when the pressure increase then density also increases at constant temperature. On the other hand Since we have relation v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$ in this relation γ is constant  and P/ρ is also constant then from the above relation we have v=a constant. At the given temperature.

Temperature:  since we have a relation v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$  we also have ρ=M/V  where M be the mass of the gas V be the volume. Then v=γPVM−−−−√$\sqrt{\frac{\gamma PV}{M}}$ ………..1

Now from ideal gas equation for one mole gas we have relation PV=RT…..2

From 1 and 2 we get

v=γRTM−−−−√$\sqrt{\frac{\gamma RT}{M}}$ since γ, R and ρ is constant then we have v=CT√……….3$C\sqrt{T}\dots \dots \dots .3$where C=γRM−−−√$\sqrt{\frac{\gamma R}{M}}$

now from equation 3 we have relation that velocity of the sound is proportional to the square root of the absolute temperature at a given pressure.

2064 Q.NO. 2 (d) REGULAR.

Explain why the velocity of the sound in solid is grater then that in gases, through the densities of solid are greater than that of gases. 

 

[Marks : 2 ] Hide Ans

Ans. :

The sound is longitudinal wave. In general the velocity of longitudinal wave is given by v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$, where γ is the modulus of elasticity and ρ is the density of the medium. For the solids, the ratio (γ/ρ) is much greater than that for gases. Hence, the velocity of sound in solids is greater than in gases, although the densities of solids are greater than that of gases.

2064 Q.NO. 5 (b) REGULAR.

A source of sound of frequency 512Hz emits wave of wavelength 670mm in air at 200C. What is the velocity of the sound in air at this temperature? What would be the wavelength of sound from the source in air at 00C? 

 

[Marks : 4 ] Hide Ans

Ans. :

Solution.

Frequency of sound (f)=512Hz

Wavelength of sound at 200C(λ20)=670mm=6.70 x 10-1m

Temperature at 200C (T20)=200C=20+273=293K

Temperature 00C(T0)=00C=0+273=273K

Velocity of sound in at 200C(V20)=?

Velocity of sound in air at 00C(V0)=?

We know, v=λf

Or, v20=λ20f= 6.70 x 10-1 x 512=343m/sec. ans.

Since the velocity is directly proportional to the square root of temperature the we have

V0V20=T0T20−−−√$\frac{V_0}{V_{20}}=\sqrt{\frac{T_0}{T_{20}}}$

or,$or,$fλ0fλ20=T0T20−−−√$\frac{f{\lambda }_0}{f{\lambda }_{20}}=\sqrt{\frac{T_0}{T_{20}}}$  where v = f λ since f is same as the medium is same

λ0=T0T20−−−√λ20${\lambda }_0=\sqrt{\frac{T_0}{T_{20}}}{\lambda }_{20}$ =273293−−−√$\sqrt{\frac{273}{293}}$ x 6.70 x 10-1=0.647m

2063 Q.NO. REGULAR.

Explain and state the stationary wave.

[Marks : 10 ] Hide Ans

Ans. :

When two wave trains of same frequency and amplitude travel with the same velocity along the same straight line in opposite directions, they superimpose and produce a new type of wave called stationary wave or standing wave.

The name stationary for such type of waves is justified because there is no flow of energy along the wave. Let the incident wave propagating along Y- axis be

25.  

and the wave reflected from the boundary traveling along negative X-axis is

25.  

when two wave superimposed then the stationery wave is produce then the resultant wave is given by y=y1 +y2=a sin(ωt+kx)+asin(ωt-kx)

or, y=sinωt Cos (-kx)=2a coskx. Sinωt

or, y=2a cos2πλ$\frac{2\pi }{\lambda }$x sinωt……..3

eq.3 represent the equation of stationary wave.

Here 2acoskx is amplitude and sinωt give the nature of the amplitude of the oscilliaration.

Special cases

If 2a coskx=0 then coskx=0, therefore x=(n+12)$\left(\mathrm{n}+\frac{1}{2}\right)$∏

Or, x=(n+12)λ2$\frac{\left(\mathrm{n}+\frac{1}{2}\right)\lambda }{2}$  where n is the interger. The point x is always at rest called node.

When n = 0

X1 = λ /4

when  n = 1

x2 = 34$\frac{3}{4}$ λ 

The difference between two different nodes x2  –   x1 =  λ/2

 

If 2a coskx =±1$\pm 1$ then, kx=n∏

Or, x=nλ/2

Here at the point x maximum displacement occur called antinodes.

2063 Q.NO. 3 (a) REGULAR.

Although te density of the solid is hight, the velocity of sound is grater in solid. Explain.

 

[Marks : 2 ] Hide Ans

Ans. :

The sound is longitudinal wave. In general the velocity of longitudinal wave is given by v=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$, where γ is the modulus of elasticity and ρ is the density of the medium. For the solids, the ratio (γ/ρ) is much greater than that for gases. Hence, the velocity of sound in solids is greater than in gases, although the densities of solids are greater than that of gases.

2062 Q.NO. 2 SUPPLEMENTARY.

State Newton’s formula for the velocity of a sound in gases. What correction was done by Laplace on it?             

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2062 Q.NO. REGULAR.

Do sound wave undergoes reflection, refraction and polarization phenomenon? Explain. 

[Marks : 4 ] Hide Ans

Ans. :

yes sound wave goes on reflection and refraction but not polarization.

1. Sound undergoes reflection as we heard echo. Eg. Reflected from wall, ground, big hall, etc.
2. Sound heard more clear in night than that of day due to refraction.
3. Sound doesn’t goes under polarization because it is mechanical wave and mechanical wave don’t follow the law of polarization but only electromagnetic wave follow law of polarization.

2062 Q.NO. REGULAR.

Which type of wave propagate in liquid, explain? 

[Marks : 4 ] Hide Ans

Ans. :

In air and gases, sound is propagated in longitudinal waves. Compression and rarefaction of gases facilitate this way of transmission. Liquids are non-compressible.

For the propagation of the transverse wave, the modulus of the rigidity of the medium is responsible,  for the propragation of the longitudinal wave bulk modulus is responsible. So, solid has longitudinal and transverse wave but liquid has only longitudinal wave. Hence in the liquid longitudinal wave is propagated. 

2062 Q.NO. 1 (c) REGULAR.

Do sound waves need a medium to travel from one point to another point in space? What properties of the medium are releveant? 

 

 

[Marks : 2 ] Hide Ans

Ans. :

Yes, sound wave needs a medium to travel from one point to other point in space because sound wave is longitudinal wave as well as mechanical wave. The relevant properties of the medium are follows:

1.The medium should possess the property of elasticity.

2.The medium should possess the property of inertia.

3.The medium should have minimum friction.

2061 Q.NO. REGULAR.

Use the principle of superposition of two waves to find the position of displacement of nodes and antinodes in a standing wave. 

[Marks : 10 ] Hide Ans

Ans. :

When two wave trains of same frequency and amplitude travel with the same velocity along the same straight line in opposite directions, they superimpose and produce a new type of wave called stationary wave or standing wave.

The name stationary for such type of waves is justified because there is no flow of energy along the wave. Let the incident wave propagating along Y- axis be

25.  

and the wave reflected from the boundary traveling along negative X-axis is

25.  

when two wave superimposed then the stationery wave is produce then the resultant wave is given by y=y1 +y2=a sin(ωt+kx)+asin(ωt-kx)

or, y=sinωt Cos (-kx)=2a coskx. Sinωt

or, y=2a cos2πλ$\frac{2\pi }{\lambda }$x sinωt……..3

eq.3 represent the equation of stationary wave.

Here 2acoskx is amplitude and sinωt give the nature of the amplitude of the oscilliaration.

Special cases

If 2a coskx=0 then coskx=0, therefore x=(n+12)$\left(\mathrm{n}+\frac{1}{2}\right)$∏

Or, x=(n+12)λ2$\frac{\left(\mathrm{n}+\frac{1}{2}\right)\lambda }{2}$  where n is the interger. The point x is always at rest called node.

When n = 0

X1 = λ /4

when  n = 1

x2 = 34$\frac{3}{4}$ λ 

The difference between two different nodes x2  –   x1 =  λ/2

 

If 2a coskx =±1$\pm 1$ then, kx=n∏

Or, x=nλ/2

Here at the point x maximum displacement occur called antinodes.

2061 Q.NO. 2 SUPPLEMENTARY.

State Newton’s formula for the velocity of a sound in gases. What correction was done by Laplace on it?             

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2060 Q.NO. 1 (c) REGULAR.

Why sound at the distance can be heard distinctly at the night than in the daytime?

 

[Marks : 2 ] Hide Ans

Ans. :

 At night, the atmospheric air becomes more cool than at the day time and reduces the density of air in the night time. As we have relation between velocity and density i.e. the velocity of sound is inversely proportional to the square root of the density (i.e. v α 1ρ√$\frac{1}{\sqrt{\rho }}$). So, at night velocity of sound is more at night than at day due to reduced density of air. On the other hand the sound absorbing bodies such as people, vehicles and noise from other sound producing bodies are least at night time as compared to the day time. So sound made at a distance can be heard more distinctly at night than at the day time.

2058 Q.NO. REGULAR.

If you are walking on the moo surface, can you hear the cracking sound behind you? Explain. 

[Marks : 4 ] Hide Ans

Ans. :

As sound required medium to travel but on the moon there is no atmosphere or it is like to be vacuum so cracking of sound is not heard behind on the moon surface.

2058 Q.NO. 5 (b) REGULAR.

At what temperature, the velocity of sound in air is increased by 50% to that at 270C? 

 

[Marks : 4 ] Hide Ans

Ans. :

Solution.

Temperature 270C(T27)=270C=27+273=300K

The velocity of sound in air at 270C(V27)= v

The velocity of sound at Tt is increase by 50% to that of 270C(Vt)= v +50% of v=1.5v

Let the be Temperature (Tt) at which the velocity increase by 50%=?

Since the velocity is directly proportional to the square root of temperature the we have

VtV27=TtT27−−−√$\frac{V_t}{V_{27}}=\sqrt{\frac{T_t}{T_{27}}}$

Squaring this equation we get

Or, = V2tV227$\frac{V_t^2}{V_{27}^2}$=TtT27$\frac{T_t}{T_{27}}$

Or, Tt=T27V2tV227$T_t=T_{27}\frac{V_t^2}{V_{27}^2}$ = 300. (1.5v/v)2=675K 

2057 Q.NO. 5 (b) REGULAR.

A man standing at one end of a closed corridor 57m long blow a short blast on a whistle. He found that the time from the blast to the sith echo was 2 second. if the temperature was 170C, what was the velocity of sound at 00C? 

 

[Marks : 4 ] Show Ans

2056 Q.NO. 1 (c) REGULAR.

Why sound are hear better on a wet day than on a dry day? 

 

 

[Marks : 2 ] Hide Ans

Ans. :

Velocity of sound is more in damp air or moisture air. We know that the density of damp air is less than air due to the presence of humidity. Also we have the relation (v α )i.e. the velocity of sound in air is inversely proportional to the square root of the density of the medium. So the velocity of sound in damp air is more than that of in dry air.

2056 Q.NO. 2 REGULAR.

Discuss Laplace correction and derive for the velocity of sound in gas. 

[Marks : 4 ] Hide Ans

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2055 Q.NO. REGULAR.

How are stationary wave formed? 

[Marks : 2 ] Hide Ans

Ans. :

When two wave progressive waves of the same wavelength and same amplitude travel through a medium in opposite direction and superimpose to each other, they give rise to wave called stationary wave.

Stationary waves have nodes when the amplitude is zero and antinodes where the amplitude is maximum. At node the particle are completely at rest but at antinode the particle are vibrating with maximum amplitude.

2055 Q.NO. 1 (b) REGULAR.

How are stationary wave formed? 

[Marks : 2 ] 

Ans. :

When two wave progressive waves of the same wavelength and same amplitude travel through a medium in opposite direction and superimpose to each other, they give rise to wave called stationary wave.

Stationary waves have nodes when the amplitude is zero and antinodes where the amplitude is maximum. At node the particle are completely at rest but at antinode the particle are vibrating with maximum amplitude.

2055 Q.NO. 2 REGULAR.

What is Newton’s formula for the velocity of sound? What correction was made by Laplace? 

[Marks : 4 ] 

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2053 Q.NO. 2 REGULAR.

What is Newton’s formula for the velocity of sound? What correction was made by Laplace? 

[Marks : 4 ] 

Ans. :

We know that the properties of a medium that govern the propagation of a mechanical wave are:

a.A restoring force

b. An inertial mass

The restoring force acting on the particles of the medium is intimately connected to the approximate elastic modulus of the medium and the inertial mass, to its density.

 Newton derived an expression for the velocity of sound in a homogenous medium is

V=Eρ−−√$\sqrt{\frac{E}{\rho }}$………………..1

where V is the velocity of sound, E is the modulus of elasticity and ρ is the density of the medium. He say that if the medium is a gas then we only considered the bulk modulus and give a relation.

V=Bρ−−√$\sqrt{\frac{B}{\rho }}$………………….2

Where B is the bulk modulus of elasticity. According to the assumed of Newton temperature remains constant when sound travels through a gas. Hence the process is isothermal and he applied Boyle’s law. At a region of compression, the pressure increases and volume decreases.

Let the initial pressure and initial volume is P and V, and the final pressure and volume are P+dP and V-dV. Here dP is increase in pressure and dV is decrease in volume at the region of compression.

Applying Boyle's law, (P + dP) (V -dV) = PV

Or, PV - dPV + dVP - dP.dV = PV…………..3

Since the changes in pressure and volume are small, dP.dV can be neglected. Then, from 3weget

- dVP + dPV =0

P=VdPdV$V\frac{dP}{dV}$…………..4

By the definition of the bulk we have B=changeinpressurechangeinvolumeoriginalvolume$\frac{changeinpressure}{\frac{changeinvolume}{originalvolume}}$

Therefore we have B=dPdVV$\frac{dP}{\frac{dV}{V}}$

Or, B=VdPdV$V\frac{dP}{dV}$………………..5

Therefore from eq. 4 and 5 we find that P=B

Therefore, Newton's formula for velocity of sound can be written as

V=Pρ−−√$V=\sqrt{\frac{P}{\rho }}$

At N TP the pressure of air P = 0.76 x 9.8 x 13.6 x 103Nm-2

= 1.013 x 105 Pa or Nm-2

Therefore V=1.013×1051.293−−−−−−−√=280m$\sqrt{\frac{1.013\times {10}^5}{1.293}}=280m$/s

But thetheoritical vaule velocity of sound at 00Cis 332ms-1.Thus Boyle's law does not apply in this case.

Latter Laplace correct the netwon formula for the velocity of sound by assume the process is adiabatic.

The relation between pressure and volume of a gas under adiabatic conditions is given by

PVγ$V^{\gamma }$=a constant.

We have γ=CPCV$\frac{C_P}{C_V}$

Let the pressure change by an amount dP, producing a change in volume by dV. Then

PVγ$V^{\gamma }$=(P+dP) (V-dV)γ

Taking out Vγ$V^{\gamma }$from the second factor from the above expession

PVγ$V^{\gamma }$=(P+dP).Vγ(1−dVV)γ${\left(1-\frac{dV}{V}\right)}^{\gamma }$…………………..6

But from the bionomial expansion

(1−dVV)γ≈1−γdVV${\left(1-\frac{dV}{V}\right)}^{\gamma }\approx 1-\gamma \frac{dV}{V}$

Now from equation 6 we have

P=(P+dP). (1−γdVV)${\left(1-\gamma \frac{dV}{V}\right)}^{}$…………………..7

P=P- γPΔVV$\frac{\mathrm{\Delta }V}{V}$ + dP-γV.dP.dV$\frac{\gamma }{V}.dP.dV$

Canceling P on both sides and neglecting the term containing dP.dV because it is too small, we get

-γPdVV$\frac{dV}{V}$ + dP=0

γPdVV$\frac{dV}{V}$ = dP

γP=VdV$\gamma P=\frac{V}{dV}$ .dP

But the LHS in the above equation represents the bulk modulus,

B= γP

From eq. 2 we have

V=γPρ−−−√$\sqrt{\frac{\gamma P}{\rho }}$

This is known as Newton-Laplace formula for the velocity of sound in a gas.

2053 Q.NO. 3 (or) REGULAR.

The interval between the flash of lighting and the sound of the thunder is 2 seconds, when the temperature is 100C. How far is the storm if the velocity of sound in air at 00C is 330m/s? 

 

[Marks : 4 ] 

Ans. :

solution 

Time taken between the interval of flash of lightening and thundered 100(t10)=2sec

Velocity at 100C(V10)= distance travel(d) /time taken=d/2

Temperature at100C (T10)=100C=10+273=283K

Temperature 00C(T0)=00C=0+273=273K

The velocity of sound in air at 00C(V0)=330m/s

Since the velocity is directly proportional to the square root of temperature the we have

V0V10=T0T10−−−√$\frac{V_0}{V_{10}}=\sqrt{\frac{T_0}{T_{10}}}$

or,$or,$330d/2=T0T10−−−√$\frac{330}{d/2}=\sqrt{\frac{T_0}{T_{10}}}$

d=2×330283273−−−√$=2\times 330\sqrt{\frac{283}{273}}$ =671.97m.

2052 Q.NO. 1 (b) REGULAR.

Difference between force vibration and free vibration. 52 1b

[Marks : 2 ] 

Ans. :

Force vibration	Free vibration
When the body vibrates with applying external force then it is called force vibration.	When the body vibrate with its own natural frequency with out the help of  external force then it is free vibration.
the vibration of a pendulum is free vibration which needs no external force to vibrate.	the vibration of a machine like a drill is forced vibration which needs an external force to vibrate.