Two closed organ pipes give $10$ beats between the fundamentals when sounded together. If the length of the shorter pipe is $1 m$ , then the length of the longer pipe will be ? (speed of sound in air is $340 m s^{- 1}$ )

2.87 m

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0.87 m

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1.13 m

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2.13 m

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Solution

The correct option is C $1.13 m$
Given, $f_{b e a t} = 10 H z, L_{A} < L_{B}, L_{A} = 1 m & v = 340 m s^{- 1}$

We know that the fundamental or first harmonic of a closed organ pipe happens at:
$f = \frac{v}{4 L}$
This implies that frequency is higher for shorter pipe A since $v$ is same.

So,
$f_{A} - f_{B} = 10$

$\frac{v}{4 L_{A}} - \frac{v}{4 L_{B}} = 10$

$\frac{340}{4 \times 1} - \frac{340}{4 L_{B}} = 10$

$85 - \frac{340}{4 L_{B}} = 10$

$\frac{340}{4 L_{B}} = 75 \Rightarrow 4 L_{B} \approx 4.53 m \Rightarrow L_{B} \approx 1.13 m$

So, C is the correct answer.

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Two closed organ pipes give 10 beats between the fundamentals when sounded together. If the length of the shorter pipe is 1 m, then the length of the longer pipe will be ? (speed of sound in air is 340 ms−1)

Two closed organ pipes give $10$ beats between the fundamentals when sounded together. If the length of the shorter pipe is $1 m$ , then the length of the longer pipe will be ? (speed of sound in air is $340 m s^{- 1}$ )