First slide

Stationary waves

Question

In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m/s, the frequency of tuning fork is (in Hz)

Easy

A

500
B

300
C

330
D

165

Solution

For closed pipe $l_{1} = \frac{v}{4 n}; l_{2} = \frac{3 v}{4 n} \Rightarrow v = 2 n (l_{2} - l_{1})$
$\Rightarrow n = \frac{v}{2 (l_{2} - l_{1})} = \frac{330}{2 \times (0 .49 - 0 .16)} = 500 Hz$

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