Mechanical Wave

Numerical

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

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

For helium

VHe = $\sqrt {\frac{{\frac{5}{3}{\rm{RT}}}}{4}}$……………..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 Vs 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 Vo 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, vo = $\sqrt {\left( {\frac{{273}}{{300}}} \right)}$ * v

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

= 333.9 m/s.

But,

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

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

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

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

= 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 vo 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 = 2vo.

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 (λo) = 1.5m

So, Speed of sound at NTP (vo) = f * λo

= 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 - λo

= 1.57 – 1.5

= 0.07m

11. Soln:

Given

Density of helium (δm) = 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 vo$\propto $$\sqrt {{{\rm{T}}_{\rm{o}}}} Speed of sound at θ°C is vo \propto$$\sqrt {\rm{T}}$

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

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

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

So, vθ = vo.

Or, vo$\sqrt {\frac{{273 + \theta }}{{273}}}$ = vθ

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

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

= 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

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

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

Tt=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= 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 Vaand that of hydrogen be Vhand d aand dhbe the densitiesofair and hydrogen

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

∴vh=4 * va=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.

Tn= 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

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