A spring of mass 0.2 kg/m has length l = 0.6 m. It is fixed at both… | MHT CET 2026 (13 May, Shift 2)

Q14MHT CET 2026 · 13 MAY · SHIFT 2Medium

Stationary Waves & Vibrating Strings

A spring of mass $0.2$ kg/m has length $l = 0.6$ m. It is fixed at both ends and stretched such that is has a tension $80$ N. The string vibrates in three segments with amplitude $0.5$ cm. The amplitude of transverse (particle) velocity is

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Answer: (c)

$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{80}{0.2}} = 20\text{ m/s}$ .
3 segments: $L = \dfrac{3\lambda}{2} \Rightarrow \lambda = 0.4\text{ m}$ , so $f = \dfrac{20}{0.4} = 50\text{ Hz}$ and $\omega = 100\pi\text{ rad/s}$ .
Maximum particle velocity (at an antinode) $= A\omega = 0.005 \times 100\pi = 0.5\pi \approx 1.57\text{ m/s}$ .
So the answer is (c) $1.57\text{ m/s}$ .

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Stationary Waves & Vibrating Strings question from MHT CET 2026 (13 May, Shift 2)

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