First slide

Stationary waves

Question

If the length of a closed organ pipe is 1m and velocity of sound is 330 m/s, then the frequency for the second note is

Moderate

A

$4 \times \frac{330}{4} Hz$
B

$3 \times \frac{330}{4} Hz$
C

$2 \times \frac{330}{4} Hz$
D

$2 \times \frac{4}{330} Hz$

Solution

For closed pipe $n_{1} = \frac{v}{4 l} = \frac{330}{4} Hz$
Second note = $3 n_{1} = \frac{3 \times 300}{4} Hz$

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