SATHEE: Wave Motion 3 Question 1

Wave Motion 3 Question 1

1. The pressure wave $p = 0.01 \sin [1000 t - 3 x] N m^{- 2}$ , corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is $0^{\circ} C$ . On some other day when temperature is $T$ , the speed of sound produced by the same blade and at the same frequency is found to be $336 m s^{- 1}$ . Approximate value of $T$ is

(2019 Main, 9 April I)

(a) $15^{\circ} C$

(b) $11^{\circ} C$

(d) $4^{\circ} C$

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Answer:

Correct Answer: 1. (d)

Solution:

Given, $p = 0.01 \sin (1000 t - 3 x) N / m^{2}$

Comparing with the general equation of pressure wave of sound, i.e. $p_{0} \sin (ω t - k x)$ ,

we get

$ω = 1000 and k = 3$

Also,

$k = \frac{ω}{v} \Rightarrow v = ω / k$

$∴$ Velocity of sound is $| v_{1} | = \frac{1000}{3}$

$O r$

Speed of sound wave can also be calculated as

$v = - \frac{(coefficient of t)}{(coefficient of x)} = - \frac{1000}{(- 3)} = \frac{1000}{3} m / s$

Now, relation between velocity of sound and temperature is

$v = \sqrt{\frac{γ R T}{m}} \Rightarrow v \propto \sqrt{T}$

or $\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} \Rightarrow T_{2} = \frac{v_{2}^{2}}{v_{1}^{2}} \cdot T_{1}$

Here, $v_{2} = 336 m / s, v_{1} = 1000 / 3 m / s$ ,

$T_{1} = 0^{\circ} C = 273 K$

$∴ T_{2} = \frac{(336)^{2}}{(1000 / 3)^{2}} \times 273 = 277.38 K$

$∴ T_{2} = {4.38}^{\circ} C ≃ 4^{\circ} C$

Wave Motion 3 Question 1

Answer:

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Wave Motion 3 Question 1

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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