Wave Motion 3 Question 1
1. The pressure wave $p=0.01 \sin [1000 t-3 x] Nm^{-2}$, 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 $336 ms^{-1}$. Approximate value of $T$ is
(2019 Main, 9 April I)
(a) $15^{\circ} C$
(b) $11^{\circ} C$
(c) $12^{\circ} C$
(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. $p _0 \sin (\omega t-k x)$,
we get
$$ \omega=1000 \text { and } k=3 $$
Also,
$$ k=\frac{\omega}{v} \Rightarrow v=\omega / k $$
$\therefore$ Velocity of sound is $\left|v _1\right|=\frac{1000}{3}$
$$ Or $$
Speed of sound wave can also be calculated as
$$ v=-\frac{(\text { coefficient of } t)}{(\operatorname{coefficient} \text { of } x)}=-\frac{1000}{(-3)}=\frac{1000}{3} m / s $$
Now, relation between velocity of sound and temperature is
$v=\sqrt{\frac{\gamma R T}{m}} \Rightarrow v \propto \sqrt{T}$
or $\quad \frac{v _2}{v _1}=\sqrt{\frac{T _2}{T _1}} \Rightarrow T _2=\frac{v _2^{2}}{v _1^{2}} \cdot T _1$
Here, $\quad v _2=336 m / s, v _1=1000 / 3 m / s$,
$$ T _1=0^{\circ} C=273 K $$
$\therefore \quad T _2=\frac{(336)^{2}}{(1000 / 3)^{2}} \times 273=277.38 K$
$\therefore T _2=4.38^{\circ} C \simeq 4^{\circ} C$
