A string of mass per unit length 0.1 kg m⁻¹ has length 0.9 m. It is… | MHT CET 2025 (19 Apr, Shift 1)

Q31MHT CET 2025 · 19 APR · SHIFT 1Medium

Stationary Waves & Vibrating Strings

A string of mass per unit length $0.1\text{ kg m}^{-1}$ has length 0.9 m. It is fixed at both ends and stretched such that it has a tension of 40 N. The string vibrates in three segments with amplitude 0.3 cm. The amplitude (maximum) of the particle velocity is (in $\text{m/s}$ )

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Answer: (c)

Wave speed $v = \sqrt{T/\mu} = \sqrt{40/0.1} = 20\text{ m/s}$ .
Three segments: $3\dfrac{\lambda}{2} = 0.9 \Rightarrow \lambda = 0.6\text{ m}$ , so $n = \dfrac{v}{\lambda} = \dfrac{100}{3}\text{ Hz}$ .
$\omega = 2\pi n = \dfrac{200\pi}{3}\text{ rad/s}$
$v_{max} = A\omega = 0.003 \times \dfrac{200\pi}{3} = 0.2\pi = \dfrac{\pi}{5}\text{ m/s}$
So the answer is (c) $\dfrac{\pi}{5}$ .

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