Physics · Transverse waves

JEE Advanced 2023 — Paper 2 — Question 23

A string of length 1 m and mass $2 \times 10^{-5} \mathrm{~kg}$ is under tension $T$ . When the string vibrates, two successive harmonics are found to occur at frequencies 750 Hz and 1000 Hz . The value of tension $T$ is $\qquad$ Newton.

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\frac{\mathrm{nv}}{2 \ell}=750 \mathrm{~Hz}$ $\frac{(\mathrm{n}+1) \mathrm{v}}{2 \ell}=1000 \mathrm{~Hz}$ $\therefore \frac{\mathrm{v}}{2 \ell}=1000-750=250 \mathrm{~Hz}$ $\Rightarrow \mathrm{v}=2 \times 250 \times 1$ $\Rightarrow \sqrt{\frac{\mathrm{T}}{\mu}}=500$ $\therefore \mathrm{T}=5 \mathrm{~N}$ .

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Physics
Chapter: Transverse waves
Topic: Standing Waves on a String and Modes of Vibration