Wave Motion 2 Question 1
1. A tuning fork of frequency $480 Hz$ is used in an experiment for measuring speed of sound $(v)$ in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column $l _1=30 cm$ and $l _2=70 cm$.
Then, $v$ is equal to
(Main 2019, 12 April II)
(a) $332 ms^{-1}$
(b) $384 ms^{-1}$
(c) $338 ms^{-1}$
(d) $379 ms^{-1}$
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Answer:
Correct Answer: 1. (b)
Solution:
- In a resonance tube apparatus, first and second resonance occur as shown
As in a stationary wave, distance between two successive nodes is $\frac{\lambda}{2}$ and distance of a node and an antinode is $\frac{\lambda}{4}$.
$$ l _2-l _1=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} $$
So, speed of sound, $v=f \lambda=f \times 2\left(l _2-l _1\right)$ $=480 \times 2 \times(70-30) \times 10^{-2}=384 ms^{-1}$
