Physics · Sound Waves

JEE Main 2024 — 1 February, Shift 1 — Question 53

A tuning fork resonates with a sonometer wire of length 1m stitched with a tension of 6N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _____ Hz.

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

f_{1} = \frac{1}{2}\sqrt{\frac{6}{\mu}}, \qquad f_{2} = \frac{1}{2}\sqrt{\frac{54}{\mu}}

\frac{f_{1}}{f_{2}} = \frac{1}{3}, \qquad f_{2} - f_{1} = 12

f_{1} = 6\,\text{Hz}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Sound Waves
Topic: Vibrations in rod and Air Columns - Organ pipes