Waves

JEE Main level, numerical answer, medium

89 of 142

Question 89 of 142

A closed organ pipe 150cm150cm long gives 77 beats per second with an open organ pipe of length 350cm350cm, both vibrating in fundamental mode. The velocity of sound is ______ ms−1m{s}^{-1}.

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Answer: 294

The formula to calculate the fundamental frequency in a closed pipe is given by
fc=v4l1...(1){f}_{c}=\dfrac{v}{4{l}_{1}}...\left(1\right)
And, that in an open pipe is given by
fo=v2l2...(2){f}_{o}=\dfrac{v}{2{l}_{2}}...\left(2\right)
Given that
∣fc−fo∣=7...(3)\left|{f}_{c}-{f}_{o}\right|=7...\left(3\right)
From equations (1), (2) and (3), it follows that
v4×150−v2×350=7⇒v600cm−v700cm=7⇒v6m−v7m=7⇒v(142)=7⇒v=42×7=294ms−1\dfrac{v}{4\times 150}-\dfrac{v}{2\times 350}=7 \Rightarrow \dfrac{v}{600cm}-\dfrac{v}{700cm}=7 \Rightarrow \dfrac{v}{6m}-\dfrac{v}{7m}=7 \Rightarrow v\left(\dfrac{1}{42}\right)=7 \Rightarrow v=42\times 7 =294m{s}^{-1}

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