A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be (a) 0.25 m (b) 0.5 m (c) 1 m (d) 2 m

Answer: (c). Same string and tension: n ∝ 1/L, so nL is constant. 512 × 0.5 = 256 × L ⇒ L = 1 m So the answer is (c) 1 m.

A stretched string resonates with tuning fork frequency 512 Hz when… | NEET 1993 (AIPMT)

Q118NEET 1993 · AIPMTMedium

Stationary Waves & Vibrating Strings

A stretched string resonates with tuning fork frequency $512\text{ Hz}$ when length of the string is $0.5\text{ m}$ . The length of the string required to vibrate resonantly with a tuning fork of frequency $256\text{ Hz}$ would be

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Answer: (c)

Same string and tension: $n \propto \dfrac{1}{L}$ , so $nL$ is constant.
$512 \times 0.5 = 256 \times L \Rightarrow L = 1\text{ m}$
So the answer is (c) $1\text{ m}$ .

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