A string fixed at both the ends forms standing wave with node separation of 5 cm. If the velocity of the wave on the string is 2 m/s, then the frequency of vibration of the string is (a) 0.2 Hz (b) 10 Hz (c) 20 Hz (d) 40 Hz

Answer: (c). Node separation (λ)/(2) = 5 cm ⇒ λ = 0.1 m. n = (v)/(λ) = 2/0.1 = 20 Hz. So the answer is (c) 20 Hz.

A string fixed at both the ends forms standing wave with node… | MHT CET 2023 (11 May, Shift 1)

Q57MHT CET 2023 · 11 MAY · SHIFT 1Easy

Stationary Waves & Vibrating Strings

A string fixed at both the ends forms standing wave with node separation of $5\text{ cm}$ . If the velocity of the wave on the string is $2\text{ m/s}$ , then the frequency of vibration of the string is

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Answer: (c)

Node separation $\frac{\lambda}{2} = 5\text{ cm} \Rightarrow \lambda = 0.1\text{ m}$ .
$n = \frac{v}{\lambda} = \frac{2}{0.1} = 20\text{ Hz}$ .
So the answer is (c) 20 Hz.

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