Calculate the wavelength of wave in air at 25^oC if the frequency of sound wave is 256 Hz and velocity of sound at 0^oC is 330 [latex]ms^{-1}[/latex].

Solution
Given,
Wavelength at 25^oC ([latex]\lambda_{25}[/latex]) =?
Frequency (f) = 256 Hz
Temperature (T₂₅) = 25^oC = 25 + 273 = 298 K
Temperature (T_o) = 0^oC = 0 + 273 = 273 K
Velocity at 0^oC ([latex]v_o[/latex]) = 330 m/s
We know that,
[latex]v \propto \sqrt{T}[/latex]

Or, [latex]\frac{v_{25}}{v_o} = \sqrt{\frac{T_{25}}{T_o}}[/latex]

Or, [latex]\frac{f\times \lambda_{25}}{v_o} = \sqrt{\frac{298}{273}}[/latex]

Or, [latex]\frac{256\times \lambda_{25}}{330} = 1.045[/latex]

Or, [latex]\lambda_{25} = \frac{330\times 1.045}{256}[/latex]

= 1.35 m/s

At what temperature, the velocity of sound in air is increased by 60% to that at 27°C?

Solution

Given,

Let, velocity at 27°C ([latex]v_{27}[/latex]) = x

Temperature ([latex]T_{27}[/latex]) = 27°C = 27 + 273 = 300 K

Velocity at T°C ([latex]v_t[/latex]) = x + 60% of x

[latex]= x + \dfrac{60}{100}\cdot x[/latex]

[latex]= 1.6x[/latex]

Temperature ([latex]T_t[/latex]) = T°C (Let) = (T + 273) K

We know that,

[latex]v \propto \sqrt{T}[/latex]

So, [latex]\dfrac{v_{27}}{v_t} = \sqrt{\dfrac{T_{27}}{T_t}}[/latex]

Or, [latex]\dfrac{x}{1.6x} = \sqrt{\dfrac{300}{T+273}}[/latex]

Or, [latex]\dfrac{1}{1.6} = \sqrt{\dfrac{300}{T+273}}[/latex]

Squaring both sides,

[latex]\dfrac{1}{2.56} = \dfrac{300}{T+273}[/latex]

Or, T + 273 = 768

Or, T = 768 – 273 = 495°C.

In a stormy day, a person observes a lightning flash which is followed by a thunder 3 s later. How would you estimate the distance of the lightning flash from the person. (Velocity of sound = 340 m/s, velocity of light = [latex]3 \times 10^8\ m/s[/latex]) Ans: 1020 m

Solution

Given,

Time interval between lightning and thunder (Δt) = 3 s

Distance of the lightning strike from the person (d) =?

Velocity of sound (v) = 340 m/s

Velocity of light (c) = [latex]3 \times 10^8\ m/s[/latex]

We know,

Time interval between lightning and thunder (Δt) = Time of thunder – Time of lightning

Or, [latex]3 = t_t – t_l[/latex]

Or, [latex]3 = \left(\dfrac{d}{v} – \dfrac{d}{c}\right)[/latex]

Or, [latex]3 = d\left(\dfrac{1}{v} – \dfrac{1}{c}\right)[/latex]

Or, [latex]3 = d\left(\dfrac{c-v}{cv}\right)[/latex]

Or, [latex]d = \dfrac{3cv}{c-v}[/latex]

[latex]= \dfrac{3 \times (3 \times 10^8) \times 340}{(3 \times 10^8 – 340)}[/latex]

[latex]= 1020\ m.[/latex]

A source of sound of frequency 550 Hz emits waves of wavelength 60 cm in air at 20°C. What would be the wavelength of sound from the source in air at 0°C?

Solution

Given,

Frequency of sound (f) = 550 Hz

Wavelength of waves ([latex]\lambda_1[/latex]) = 60 cm = 0.6 m

Temperature ([latex]T_1[/latex]) = 20°C = 20 + 273 = 293 K

Wavelength of sound at 0°C ([latex]\lambda_2[/latex]) =?

Temperature ([latex]T_2[/latex]) = 0°C = 0 + 273 = 273 K

We know that,

Velocity of sound at 20°C ([latex]v_1[/latex]) = [latex]f \times \lambda_1[/latex]

[latex]= 550 \times 0.6 = 330\ m/s[/latex]

Again, we have,

[latex]v \propto \sqrt{T}[/latex]

Or, [latex]\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{330}{v_2} = \sqrt{\dfrac{293}{273}}[/latex]

Or, [latex]\dfrac{330}{v_2} = 1.036[/latex]

Or, [latex]v_2 = \dfrac{330}{1.036} = 318.533\ m/s[/latex]

Again, [latex]v_2 = f \times \lambda_2[/latex]

Or, [latex]\lambda_2 = \dfrac{v_2}{f} = \dfrac{318.533}{550} = 0.579\ m = 579\ mm.[/latex]

What is the difference between the speed of longitudinal waves in air at 27°C and at –13°C? What is the speed at 0°C?

Solution

Given,

Temperature ([latex]T_1[/latex]) = 27°C = 27 + 273 = 300 K

Temperature ([latex]T_2[/latex]) = -13°C = -13 + 273 = 260 K

Difference between the speed of longitudinal waves in air at [latex]T_1[/latex] and [latex]T_2[/latex] (Δv) =?

Case I:

We know that, speed of sound in air at 0°C

[latex]v_3 = \sqrt{\dfrac{\gamma P}{\rho}}[/latex]

[latex]= \sqrt{\dfrac{1.4 \times (1.01 \times 10^5)}{1.293}}[/latex]

[latex]= 330.693\ m/s[/latex]

Case II:

We know, [latex]v \propto \sqrt{T}[/latex]

Or, [latex]\dfrac{v_1}{v_3} = \sqrt{\dfrac{T_1}{T_3}}[/latex]

Or, [latex]\dfrac{v_1}{330.693} = \sqrt{\dfrac{300}{273}}[/latex]

Or, [latex]v_1 = 1.048 \times 330.693[/latex]

[latex]= 346.566\ m/s[/latex]

Case III:

[latex]\dfrac{v_2}{v_3} = \sqrt{\dfrac{T_2}{T_3}}[/latex]

Or, [latex]\dfrac{v_2}{330.693} = \sqrt{\dfrac{260}{273}}[/latex]

Or, [latex]v_2 = 0.976 \times 330.693[/latex]

[latex]= 322.756\ m/s[/latex]

So, Difference between the speed of longitudinal waves in air at [latex]T_1[/latex] and [latex]T_2[/latex] (Δv) = [latex]v_1 – v_2[/latex]

[latex]= 346.566 – 322.756[/latex]

[latex]= 23.81\ m/s.[/latex]

At what temperature, the velocity of sound in air is increased by 50% to that at 27°C?

Solution

Given,

Let, velocity at 27°C ([latex]v_{27}[/latex]) = x

Temperature ([latex]T_{27}[/latex]) = 27°C = 27 + 273 = 300 K

Velocity at T°C ([latex]v_t[/latex]) = x + 50% of x

[latex]= x + \dfrac{50}{100}\cdot x[/latex]

[latex]= 1.5x[/latex]

Temperature ([latex]T_t[/latex]) = T°C (Let) = (T + 273) K

We know that,

[latex]v \propto \sqrt{T}[/latex]

So, [latex]\dfrac{v_{27}}{v_t} = \sqrt{\dfrac{T_{27}}{T_t}}[/latex]

Or, [latex]\dfrac{x}{1.5x} = \sqrt{\dfrac{300}{T+273}}[/latex]

Or, [latex]\dfrac{1}{1.5} = \sqrt{\dfrac{300}{T+273}}[/latex]

Squaring both sides,

[latex]\dfrac{1}{2.25} = \dfrac{300}{T+273}[/latex]

Or, T + 273 = 675

Or, T = 675 – 273 = 402°C.

In a resonance air column apparatus, the first and second resonance positions were observed at 18 cm and 56 cm respectively. The frequency of tuning fork used was 480 Hz. Calculate the velocity of sound in air and end correction of the tube.

Solution

Given,

First resonance position ([latex]l_1[/latex]) = 18 cm = 0.18 m

Second resonance position ([latex]l_2[/latex]) = 56 cm = 0.56 m

Frequency of tuning fork (f) = 480 Hz

Velocity of sound in air (v) =?

End correction of the tube (e) =?

We know that,

[latex]v = 2f(l_2 – l_1)[/latex]

[latex]= 2 \times 480(0.56 – 0.18)[/latex]

[latex]= 364.8\ m/s[/latex]

Again, end correction (e) = [latex]\dfrac{l_2 – 3l_1}{2}[/latex]

[latex]= \dfrac{0.56 – 3 \times 0.18}{2}[/latex]

[latex]= 0.01\ m.[/latex]

Calculate the bulk modulus of a liquid in which longitudinal waves with frequency of 250 Hz have the wavelength of 8 m if the density of liquid is 900 [latex]kgm^{-3}[/latex].

Solution

Given,

Bulk modulus of a liquid (B) =?

Frequency of wave (f) = 250 Hz

Wavelength (λ) = 8 m

Density of liquid (ρ) = 900 [latex]kgm^{-3}[/latex]

We know that,

Velocity of sound in liquid (v) = [latex]f \times \lambda[/latex]

[latex]= 250 \times 8[/latex]

[latex]= 2000\ m/s[/latex]

Again,

Velocity of sound in liquid (v) = [latex]\sqrt{\dfrac{B}{\rho}}[/latex]

Or, [latex]2000 = \sqrt{\dfrac{B}{900}}[/latex]

Or, [latex]4000000 = \dfrac{B}{900}[/latex]

Or, [latex]B = 3.6 \times 10^9\ N/m^2.[/latex]

A source of sound produces a note of 512 Hz in air at 17°C with wavelength 66.5 cm. Find the ratio of molar heat capacities at constant pressure to constant volume at NTP. (Densities of air and mercury at NTP are 1.293 kg/m³ and 13600 kg/m³ respectively).

Solution

Given,

Frequency of sound (f) = 512 Hz

Temperature ([latex]T_1[/latex]) = 17°C = 17 + 273 = 290 K

Wavelength ([latex]\lambda_1[/latex]) = 66.5 cm = 0.665 m

Velocity of sound at [latex]T_1[/latex] ([latex]v_1[/latex]) = [latex]f \times \lambda_1[/latex]

[latex]= 512 \times 0.665 = 340.48\ m/s[/latex]

Now, ratio of molar heat capacity at constant pressure to constant volume at NTP (γ) =?

Density of air ([latex]\rho_a[/latex]) = 1.293 kg/m³

Density of mercury ([latex]\rho_m[/latex]) = 13600 kg/m³

We know that, atmospheric pressure (P) = [latex]h\rho_m g[/latex]

[latex]= 0.76 \times 13600 \times 9.8[/latex] [∵ height of mercury column = 760 mm = 0.76 m]

[latex]= 1.01 \times 10^5\ N/m^2[/latex]

Again,

Velocity of sound in air at NTP (0°C):

[latex]\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{340.48}{v_2} = \sqrt{\dfrac{290}{273}}[/latex]

Or, [latex]\dfrac{340.48}{v_2} = 1.031[/latex]

Or, [latex]v_2 = \dfrac{340.48}{1.031} = 330.242\ m/s[/latex]

Now, velocity of sound in air at NTP ([latex]v_2[/latex]) = [latex]\sqrt{\dfrac{\gamma P}{\rho_a}}[/latex]

Or, [latex]330.242 = \sqrt{\dfrac{\gamma \times (1.01 \times 10^5)}{1.293}}[/latex]

Or, [latex]1.091 \times 10^5 = \gamma \times (7.811 \times 10^4)[/latex]

Or, [latex]\gamma = 1.397 \approx 1.4.[/latex]

A source of sound of frequency 512 Hz emits waves of wavelength 64.5 cm in air at 20°C. What would be the velocity of sound at 0°C?

Solution

Given,

Frequency of sound (f) = 512 Hz

Wavelength ([latex]\lambda_1[/latex]) = 64.5 cm = 0.645 m

Temperature ([latex]T_1[/latex]) = 20°C = 20 + 273 = 293 K

So, velocity of sound at [latex]T_1[/latex] ([latex]v_1[/latex]) = [latex]f \times \lambda_1[/latex]

[latex]= 512 \times 0.645[/latex]

[latex]= 330.24\ m/s[/latex]

Temperature ([latex]T_2[/latex]) = 0°C = 0 + 273 = 273 K

Velocity of sound at [latex]T_2[/latex] ([latex]v_2[/latex]) =?

We know that,

[latex]\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{330.24}{v_2} = \sqrt{\dfrac{293}{273}}[/latex]

Or, [latex]\dfrac{330.24}{v_2} = 1.036[/latex]

Or, [latex]v_2 = 318.764\ m/s.[/latex]

In a stormy day, a boy observes a lighting flash which is followed by a thunder 3 secs. later. How would you estimate the distance of the lighting strike from the boy? (given velocity of sound on that day = 332 m/s, velocity of light c = [latex]3 \times 10^8\ m/s[/latex]).

Solution

Given,

Time interval between lighting and thunder (Δt) = 3 s

Distance of the lighting strike from the boy (d) =?

Velocity of sound (v) = 332 m/s

Velocity of light (c) = [latex]3 \times 10^8\ m/s[/latex]

We know,

Time interval between lighting and thunder (Δt) = Time of thunder – Time of lighting

Or, [latex]3 = t_t – t_l[/latex]

Or, [latex]3 = \left(\dfrac{d}{v} – \dfrac{d}{c}\right)[/latex]

Or, [latex]3 = d\left(\dfrac{1}{v} – \dfrac{1}{c}\right)[/latex]

Or, [latex]3 = d\left(\dfrac{c-v}{cv}\right)[/latex]

Or, [latex]d = \dfrac{3cv}{c-v}[/latex]

[latex]= \dfrac{3 \times (3 \times 10^8) \times 332}{((3 \times 10^8) – 332)}[/latex]

[latex]= 996\ m.[/latex]

When a detonator is exploded on a railway line, an observer standing on the rail 2 km away hears two sounds. What is the time interval between them? (Young’s modulus of steel = [latex]2 \times 10^{11}\ Nm^{-2}[/latex], density of steel = [latex]8 \times 10^3\ kgm^{-3}[/latex], density of air = 1.4 [latex]kgm^{-3}[/latex], γ − for air = 1.4, atmospheric pressure = [latex]10^5\ Nm^{-2}[/latex]).

Solution

Given,

Distance between observer and explosion (d) = 2 km = 2000 m

Time interval between two sounds (Δt) =?

Young’s modulus of steel ([latex]Y_{steel}[/latex]) = [latex]2 \times 10^{11}\ Nm^{-2}[/latex]

Density of steel ([latex]\rho_s[/latex]) = [latex]8 \times 10^3\ kg/m^3[/latex]

Density of air ([latex]\rho_a[/latex]) = 1.4 kg/m³

γ − for air = 1.4

Atmospheric pressure (P) = [latex]10^5\ N/m^2[/latex]

We know that,

Velocity of sound in steel ([latex]v_s[/latex]) = [latex]\sqrt{\dfrac{Y_{steel}}{\rho_s}}[/latex]

[latex]= \sqrt{\dfrac{2 \times 10^{11}}{8 \times 10^3}} = 5000\ m/s[/latex]

Time taken by sound to reach observer through steel ([latex]t_s[/latex]) = [latex]\dfrac{d}{v_s} = \dfrac{2000}{5000} = 0.4\ s[/latex]

Again,

Velocity of sound in air ([latex]v_a[/latex]) = [latex]\sqrt{\dfrac{\gamma P}{\rho_a}}[/latex]

[latex]= \sqrt{\dfrac{1.4 \times 10^5}{1.4}} = 316.228\ m/s[/latex]

Now, time taken by sound to reach observer through air ([latex]t_a[/latex]) = [latex]\dfrac{d}{v_a} = \dfrac{2000}{316.228} = 6.324\ s[/latex]

So, Time interval between two sounds (Δt) = [latex]t_a – t_s[/latex]

[latex]= 6.324 – 0.4[/latex]

[latex]= 5.924\ s.[/latex]

A source of sound of frequency 512 Hz emits waves of wavelength 670 mm in air at 20°C. What is the velocity of sound in air at this temperature? What would be the wavelength of sound from the source in air at 0°C?

Solution

Given,

Frequency of sound (f) = 512 Hz

Wavelength of waves ([latex]\lambda_1[/latex]) = 670 mm = 0.67 m

Temperature ([latex]T_1[/latex]) = 20°C = 20 + 273 = 293 K

Velocity of sound in air at [latex]T_1[/latex] ([latex]v_1[/latex]) =?

Wavelength of sound at 0°C ([latex]\lambda_2[/latex]) =?

Temperature ([latex]T_2[/latex]) = 0°C = 0 + 273 = 273 K

We know that,

Velocity of sound at [latex]T_1[/latex] ([latex]v_1[/latex]) = [latex]f \times \lambda_1[/latex]

[latex]= 512 \times 0.67 = 343.04\ m/s[/latex]

Again, we have,

[latex]v \propto \sqrt{T}[/latex]

Or, [latex]\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}[/latex]

Or, [latex]\dfrac{343.04}{v_2} = \sqrt{\dfrac{293}{273}}[/latex]

Or, [latex]\dfrac{343.04}{v_2} = 1.036[/latex]

Or, [latex]v_2 = \dfrac{343.04}{1.036} = 331.12\ m/s[/latex]

Again,

[latex]v_2 = f \times \lambda_2[/latex]

Or, [latex]\lambda_2 = \dfrac{v_2}{f} = \dfrac{331.12}{512} = 0.6467\ m = 646.7\ mm.[/latex]

A man standing at one end of a closed corridor 57 m long blow a short blast on a whistle. He found that the time from the blast to the sixth echo was 2 seconds. If the temperature was 17°C, what was the velocity of sound at 0°C?

Solution

Given,

Length of closed corridor (d) = 57 m

For the first echo, the distance covered by the sound is [latex]2 \times 57 = 114\ m[/latex].

So, the distance covered by the sound in sixth echo = [latex]6 \times 114 = 684\ m[/latex].

Now, time of sixth echo (t) = 2 seconds.

Temperature ([latex]T_{17}[/latex]) = 17°C = 17 + 273 = 290 K

Velocity of sound at 0°C = ?

Temperature ([latex]T_0[/latex]) = 0°C = 0 + 273 = 273 K

We know, velocity of sound at 17°C = [latex]\dfrac{\text{distance covered by sound at }17°C}{\text{time taken}}[/latex]

[latex]v_{17} = \dfrac{684}{2} = 342\ m/s[/latex]

Again, we know,

[latex]v \propto \sqrt{T}[/latex]

So, [latex]\dfrac{v_{17}}{v_0} = \sqrt{\dfrac{T_{17}}{T_0}}[/latex]

Or, [latex]\dfrac{342}{v_0} = \sqrt{\dfrac{290}{273}}[/latex]

Or, [latex]v_0 = 331.82\ m/s.[/latex]

The interval between the flash of lighting and the sound of thunder is 2 seconds, when temperature is 10°C. How far is the storm if the velocity of sound in air at 0°C is 330 [latex]ms^{-1}[/latex]?

Solution

Given,

Time interval between lighting and sound of thunder (t) = [latex]t_s – t_l[/latex] = 2 seconds.

Temperature ([latex]T_{10}[/latex]) = 10°C = 10 + 273 = 283 K

Distance of the storm (d) = ?

Velocity of sound in air at 0°C ([latex]V_0[/latex]) = 330 m/s

Temperature ([latex]T_0[/latex]) = 0°C = 0 + 273 = 273 K

We know that,

[latex]v \propto \sqrt{T}[/latex]

So, [latex]\dfrac{v_{10}}{v_0} = \sqrt{\dfrac{T_{10}}{T_0}}[/latex]

Or, [latex]\dfrac{v_{10}}{330} = \sqrt{\dfrac{283}{273}}[/latex]

Or, [latex]v_{10} = 335.98\ m/s[/latex]

Now, we know, distance (d) = v . t = [latex]335.98 \times 2 = 672\ m.[/latex]