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Waves

JEE Main level, single correct, medium

78 of 142

Question 78 of 142

A string is stretched between fixed points separated by $75 \mathrm{~cm}$ . It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$ . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Show answer and solution

Answer: (B) $105 \mathrm{~Hz}$

\frac{\mathrm{n}}{2 \ell}(\mathrm{v})=315, \frac{(\mathrm{n}+1)}{2 \ell} \mathrm{v}=420

Solving

\frac{\mathrm{v}}{2 \ell}=105

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Waves (Vibration of String), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.