1. Soln:

Givne:

The wavelength of the emitted note (λ) = 1m

The velocity of sound I air (v) = 340m/s.

Frequency of the note (f) = ?

Now,

Or, f = $\frac{{\rm{v}}}{\lambda }$

Or, f = $\frac{{340}}{1}$

= 340 Hz.

 

2. Soln:

Given:

CaseI: The velocity of sound in air (v) = 340m/s

Frequency of sound (f) = 256Hz

Wavelength of the sound (λ) = ?

We know,

Or, λ = $\frac{{\rm{v}}}{{\rm{f}}}$

= $\frac{{340}}{{256}}$

= 1.33m

Case II:

The velocity of sound in air (v) = 340m/s

Frequency of sound (f) = ?

Wavelength of the sound (λ) = 0.85m

We know,

Or, λ = $\frac{{\rm{v}}}{{\rm{f}}}$

Or, f = $\frac{{\rm{v}}}{\lambda }$

= $\frac{{340}}{{0.85}}$

= 400Hz.

 

3.

The speed of sound at any temperature is given

 V = $\sqrt {\frac{{\gamma {\rm{RT}}}}{{\rm{M}}}} $

 

 For hydrogen

V_h = $\sqrt {\frac{{\frac{7}{5}{\rm{RT}}}}{2}} $ ……………..i

For helium

V_He = $\sqrt {\frac{{\frac{5}{3}{\rm{RT}}}}{4}} $……………..ii

 

$\frac{{{{\rm{v}}_{\rm{h}}}}}{{{{\rm{v}}_{{\rm{he}}}}}}{\rm{\: }} = {\rm{\: }}\sqrt {\frac{{84}}{{50}}{\rm{\: }}} $=1.29

 

4. Soln:

Given normal pressure = 0.76 * 13600 * 9.8 N/m²

Or, γ = $\frac{{{{\rm{C}}_{\rm{p}}}}}{{{{\rm{C}}_{\rm{v}}}}} = \frac{{1.02}}{{0.72}}$

= 1.42

Velocity of sound at 0°C.

= $\sqrt {\frac{{\gamma {\rm{P}}}}{{\rm{r}}}} $

= $\sqrt {\frac{{1.42{\rm{*}}0.76{\rm{*}}13600{\rm{*}}9.8}}{{1.29}}} $

= 333.9m/s.

Let V_s be the velocity of sound at 27°C, then.

Or, $\frac{{{{\rm{V}}_1}}}{{{{\rm{V}}_{\rm{o}}}}} = \sqrt {\frac{{273 + 27}}{{273}}} $

= 350m/s

 

5. Soln:

Let V_o be the velocity of sound at 0°C and v be the velocity of sound at 30°C, then,

Or, $\frac{{{{\rm{v}}_{\rm{o}}}}}{{\rm{v}}} = \sqrt {\left( {\frac{{273}}{{273 + 27}}} \right)} $

Or, v_o = $\sqrt {\left( {\frac{{273}}{{300}}} \right)} $ * v

= $\sqrt {\frac{{273}}{{300}}} $ * 350

= 333.9 m/s.

But,

v_o= $\sqrt {\frac{{\gamma {\rm{p}}}}{{\rm{r}}}} $

= $\sqrt {\frac{{\gamma {\rm{*}}0.76{\rm{*}}13600{\rm{*}}9.8}}{{1.29}}} $

Or, $\frac{{\gamma {\rm{*}}0.76{\rm{*}}13600{\rm{*}}9.8}}{{1.29}} = {\rm{v}}_{\rm{o}}^2 = {\left( {333.9} \right)^2}$

Or, γ = $\frac{{1.29{\rm{*}}{{\left( {333.9} \right)}^2}}}{{0.76{\rm{*}}13600{\rm{*}}9.8}}$

= 1.42.

 

6. Soln:

Given,

At 20°C,

The wavelength (λ) = 600mm = 0.6m

The frequency (f) = 500Hz.

Velocity (v) = ?

We know,

v = fλ

= 550 * 0.6

= 330m/s.

Let v_o be the velocity of sound at 0°C, then,

Or, $\frac{{{{\rm{v}}_{\rm{o}}}}}{{\rm{v}}} = \sqrt {\left( {\frac{{273}}{{273 + 20}}} \right)} $

Or, v_o = $\sqrt {\frac{{273}}{{293}}} $ * v

Or, v_o = $\sqrt {\frac{{273}}{{293}}} {\rm{*}}330$

= 318.54 m/s.

So, Wavelength of sound at 0°C,

Or, λ = $\frac{{{{\rm{v}}_{\rm{o}}}}}{{\rm{f}}}$

= $\frac{{318.54}}{{550}}$

= 0.579m

= 579mm

 

7. Soln:

Given: The velocity of sound doesn’t change with the change in pressure, only the change in temperature is considered

T₁ = 273 + 20 = 293K

T₂ = 293 + 20 = 313K

v₁ = 340m/s.

v₂ = ?

We know,

Or, $\frac{{{{\rm{v}}_2}}}{{{{\rm{v}}_1}}} = \sqrt {\frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_1}}}} $

= $\sqrt {\frac{{313}}{{293}}} $

Or, v₂ = $\sqrt {\frac{{313}}{{293}}} {\rm{*}}{{\rm{V}}_1}$

= $\sqrt {\frac{{313}}{{293}}} {\rm{*}}340$

= 351.4 m/s.

 

8. Soln:

We know, Molar mass of hydrogen 2g = 2 * 10^-3 kg

T = 273K

R = 8.31J mol^-1 K^-1.

So, V = $\sqrt {\frac{{\gamma {\rm{RT}}}}{{\rm{M}}}} $

= $\sqrt {\frac{{1.4{\rm{*}}8.31{\rm{*}}273}}{{2{\rm{*}}{{10}^{ - 3}}}}} $

= 1262.2m/s.

 

9. Soln:

Let v_o be the velocity of sound at 0°C and t°C be the temperature at which the velocity of sound will be double the velocity of sound in air of at 0°C.

I.e. v₁ = 2v_o.

So, $\frac{{{{\rm{v}}_1}}}{{{{\rm{v}}_{\rm{o}}}}} = \sqrt {\frac{{273 + {\rm{t}}}}{{273}}} $

Or, $\frac{{2{{\rm{v}}_{\rm{o}}}}}{{{{\rm{v}}_{\rm{o}}}}} = \sqrt {\frac{{273 + {\rm{t}}}}{{273}}} $

Or, 2 = $\sqrt {\frac{{273 + {\rm{t}}}}{{273}}} $

Squaring both sides,

Or, 4 = $\frac{{273 + {\rm{t}}}}{{273}}$

Or, 4 * 273 = 273 + t

Or, t = 4 * 273 – 273

= 819°C

So, 819 + 273

= 1092K.

 

10. SolnL

The frequency of sound (f) = 220Hz.

Wavelength of sound at 0°C (λ_o) = 1.5m

So, Speed of sound at NTP (v_o) = f * λ_o

= 220 * 1.5

= 330m/s.

Now, speed of sound at 27°C is v

So, $\frac{{\rm{v}}}{{{{\rm{v}}_{\rm{o}}}}} = \sqrt {\frac{{273 + 27}}{{273}}} $

Or, v = 330 * $\sqrt {\frac{{300}}{{273}}} $

Again,

v = f * λ

So, λ = $\frac{{\rm{v}}}{{\rm{f}}}$

= $\frac{{330}}{{220}}{\rm{*}}\sqrt {\frac{{330}}{{273}}} $

= 1.57m

So, Increase in wavelength $\Delta \lambda $ = λ - λ_o

= 1.57 – 1.5

= 0.07m

 

11. Soln:

Given

Density of helium (δ_m) = 2δh

Let θ°C be the required temperature at which speed of sound in hydrogen is the same as that in helium at 0°C (at NTP).

For hydrogen,

Speed of sound at 0°C is v_o$ \propto $$\sqrt {{{\rm{T}}_{\rm{o}}}} $

Speed of sound at θ°C is v_o$ \propto $$\sqrt {\rm{T}} $

So, $\frac{{{{\rm{v}}_\theta }}}{{{{\rm{v}}_{\rm{o}}}}} = \sqrt {\frac{{\rm{T}}}{{{{\rm{T}}_{\rm{o}}}}}} $

Or, v_θ = v_o.$\sqrt {\frac{{\rm{T}}}{{{{\rm{T}}_{\rm{o}}}}}} $ = v_o$\sqrt {\frac{{273 + \theta }}{{273}}} $….(i)

Let the speed of speed of sound in helium at 0°C is v_o.

So, v_θ = v_o.

Or, v_o$\sqrt {\frac{{273 + \theta }}{{273}}} $ = v_θ

Or, $\frac{{{{\rm{v}}_\theta }}}{{{{\rm{v}}_{\rm{o}}}}} = \sqrt {\frac{{273 + \theta }}{{273}}} $

Now,

Or, $\frac{{{{\rm{v}}_\theta }}}{{{{\rm{v}}_{\rm{o}}}}} = \sqrt {\frac{{{\delta _{\rm{h}}}}}{{{\delta _{{\rm{he}}}}}}}  = \sqrt {\frac{1}{2}} $

So, $\sqrt {\frac{{273 + \theta }}{{273}}}  = \sqrt {\frac{1}{2}} $

Or, $\frac{{273 + \theta }}{{273}} = \frac{1}{2}$

So, θ = 136.5°C

 

12.

Given,

At 20°C,

The wavelength (λ) = 600mm = 0.6m

The frequency (f) = 500Hz.

Velocity (v) = ?

We know,

v = fλ

= 550 * 0.6

= 330m/s.

Let v_o be the velocity of sound at 0°C, then,

Or, $\frac{{{{\rm{v}}_{\rm{o}}}}}{{\rm{v}}} = \sqrt {\left( {\frac{{273}}{{273 + 20}}} \right)} $

Or, v_o = $\sqrt {\frac{{273}}{{293}}} $ * v

Or, v_o = $\sqrt {\frac{{273}}{{293}}} {\rm{*}}330$

= 318.54 m/s.

So, Wavelength of sound at 0°C,

Or, λ = $\frac{{{{\rm{v}}_{\rm{o}}}}}{{\rm{f}}}$

= $\frac{{318.54}}{{550}}$

= 0.579m

= 579mm

13.

14.

 

15.

Soln:

Given

Lett₁ be the time takenby the ball to reach into the well .v is the velocity of the sound

t₂is thetime taken by the soundto arriveat the surface of the well.

 Total timet = t1 + t2= 4.23s

For downward motion

 S= ut+ ½ g t ₁²

u=0, 78.4 =½ *9.8* t ₁²

t₁= 4s

for upward motion

t₂₌t- t ₁= 4.23-4 =0.23 s

 Velocity of sound in air v = $\frac{{\rm{s}}}{{\rm{t}}}{\rm{\: }} = \frac{{78.4}}{{0.23}}$= 340.86m/s

 

16.

 Let at ${{\rm{T}}_{\rm{t}}}$temperature speed of the sound is doubled

$\frac{{{{\rm{v}}_{\rm{t}}}}}{{{{\rm{v}}_0}}}$= $\sqrt {\frac{{{{\rm{T}}_{\rm{t}}}}}{{{{\rm{T}}_0}}}{\rm{\: }}} $

 

$\frac{{2{{\rm{v}}_0}}}{{{{\rm{v}}_0}}}$= $\sqrt {\frac{{{{\rm{T}}_{\rm{t}}}}}{{273}}{\rm{\: }}} $

 

T_t=4 * 273=1092 K=819° c

 

17.

Velocity at 11 ° C or 284K = v

Let at ${{\rm{T}}_{\rm{t}}}$temperature at which speed of the sound is 2/3 times of v

$\frac{{{{\rm{v}}_{\rm{t}}}}}{{{{\rm{v}}_{284}}}}$= $\sqrt {\frac{{{{\rm{T}}_{\rm{t}}}}}{{284}}{\rm{\: }}} $

$\frac{{\frac{2}{3}{\rm{v}}}}{{\rm{V}}}$= $\sqrt {\frac{{{{\rm{T}}_{\rm{t}}}}}{{284}}{\rm{\: }}} $

                $\frac{4}{3} = \frac{{{{\rm{T}}_{\rm{t}}}}}{{284}}$

 

T­_t= 378.66K

 

18.

 

 

19.

The speed of sound at anygasis given

 V = $\sqrt {\frac{{\gamma {\rm{P}}}}{{\rm{d}}}} $

 

 At STP the speed of the sound is V_aand that of hydrogen be V_hand d _aand d_hbe the densitiesofair and hydrogen

We have$\frac{{{{\rm{v}}_{\rm{a}}}}}{{{{\rm{v}}_{\rm{h}}}}}{\rm{\: }} = {\rm{\: }}\sqrt {\frac{{{{\rm{d}}_{\rm{h}}}}}{{{{\rm{d}}_{\rm{a}}}}}} $=$\sqrt {\frac{1}{{16}}} $= $\frac{1}{4}$

 

∴v_h=4 * v_a=4*332= 1328m/s

 

 

 

20.

Soln:

Given

 Lett₁ be the time takenby the stoneto reach into the water .v is the velocity of the sound

t₂is thetime taken by the soundto arriveat top of tower

for upward motion

Velocity of sound in air v}} = $\frac{{\rm{s}}}{{{{\rm{t}}_2}}}{\rm{\: }}$

$340{\rm{\: }} = \frac{{300{\rm{\: }}}}{{{{\rm{t}}_2}}}$

${{\rm{t}}_2}{\rm{\: }} = 0.88{\rm{s\: }}$

For downward motion

 S= ut+ ½ g t ₁²

u=0, 300 =½ *9.8* t ₁²

t₁= 7.82s

 

 Total timet = t1 + t2= 8.7s

 

21.

T_n= 14+273 = 287K

The speed of sound at any temperature is given

 V = $\sqrt {\frac{{\gamma {\rm{RT}}}}{{\rm{M}}}} $

 The speed of sound in oxygen is equal to the speed of the sound in nitrogen at 287 K

Now

$\sqrt {\frac{{\gamma {\rm{R}}\left( {273 + {\rm{t}}} \right)}}{{{{\rm{M}}_{\rm{o}}}}}} $=$\sqrt {\frac{{\gamma {\rm{R\: *\: }}287}}{{{{\rm{M}}_{\rm{n}}}}}} $

273 + t=$\frac{{{{\rm{m}}_0}}}{{{{\rm{M}}_{\rm{n}}}}}$* 287

t= $\frac{{16}}{{14}}{\rm{*}}287{\rm{\: }} - {\rm{\: }}273$= 55° C