Physics · Sound Waves

JEE Main 2024 — 5 April, Shift 2 — Question 53

A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with fundamental frequency of 600 Hz under same tension $\qquad$ cm .

Answer: 60

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{f}_{0}=400 \mathrm{~Hz} ; \quad \mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}=$ constant

$\frac{\lambda}{2}=\mathrm{L} ; \quad \mathrm{v}=\mathrm{f}_{0} \lambda$

$\frac{\mathrm{v}}{2 \mathrm{f}_{0}}=\mathrm{L} \Rightarrow \mathrm{v}=2 \mathrm{Lf}_{0}$

$L^{\prime}=\frac{v}{2 f^{\prime}}=\frac{2 L f_{0}}{2 f^{\prime}}$

$=\frac{\mathrm{Lf}_{0}}{\mathrm{f}^{\prime}}=\frac{90 \times 400}{600}=60$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Physics
Chapter: Sound Waves
Topic: Speed, Intensity and Characteristics of Sound Waves