JEE Main 2023 · Physics · Waves and Sound · medium

A sonometer wire vibrates in its fundamental mode at 300 Hz. Its tension is now increased by 2% and, at the same time, the vibrating length is…

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A sonometer wire vibrates in its fundamental mode at

300\ \mathrm{Hz}

. Its tension is now increased by

2\%

and, at the same time, the vibrating length is reduced by

1\%

. The number of beats per second (approximately) heard between the old and the new fundamental is

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Answer: (C) $6$

f=\dfrac{1}{2L}\sqrt{T/\mu}

, so

\dfrac{\Delta f}{f}\approx\dfrac12\dfrac{\Delta T}{T}-\dfrac{\Delta L}{L}=\tfrac12(2\%)-(-1\%)=2\%

. Hence

\Delta f\approx0.02\times300=6\ \mathrm{Hz}

, and the beat frequency equals this difference: about

6

beats per second (exact calculation:

300(\sqrt{1.02}/0.99-1)=6.1

). Forgetting that a shorter length raises the frequency gives

3

Original question written in the JEE Main 2023 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake