JEE Main 2017 · Physics · Simple Harmonic Motion · medium

Two identical sonometer wires, stretched with equal tensions, are in unison. The tension in one wire is then increased by 4% and 4 beats per second…

Single correct option

Two identical sonometer wires, stretched with equal tensions, are in unison. The tension in one wire is then increased by

4\%

and

4

beats per second are heard. The original fundamental frequency of each wire is nearly

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Answer: (A) $200\ \mathrm{Hz}$

The fundamental frequency of a stretched wire is

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

, so

f \propto \sqrt T

for a fixed wire. A

4\%

rise in tension gives

f' = f\sqrt{1.04} \approx 1.0198f

, i.e. a rise of about

2\%

(half the fractional change in

T

). The beat frequency is

f' - f \approx 0.02f = 4\ \mathrm{Hz}

, hence

f \approx 200\ \mathrm{Hz}

(exactly

202\ \mathrm{Hz}

). A common slip is to take the fractional change in frequency as

4\%

, which gives

100\ \mathrm{Hz}

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake