JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Waves & Sound

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A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. 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

Solution

Frequency of a stretched string:

f = \frac{1}{2 L} \frac{T}{μ}

, so

f \propto T

\frac{f _{2}}{f _{1}} = \frac{54}{6} = 9 = 3 ⟹ f_{2} = 3 f_{1}

Given

f_{2} - f_{1} = 12

(beats per second, with

f_{2} > f_{1}

3 f_{1} - f_{1} = 12 ⟹ f_{1} = 6 Hz

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer

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