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NEET 2020 (Phase 2) · Stationary Waves & Vibrating Strings

Stationary Waves & Vibrating Strings question from NEET 2020 (Phase 2)

Q81NEET 2020 · PHASE 2Easy

Stationary Waves & Vibrating Strings

The length of the string of a musical instrument is 90 cm and has a fundamental frequency of 120 Hz. Where should it be pressed to produce fundamental frequency of 180 Hz?

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Answer: (c)

Same tension and string, so $f \propto \dfrac{1}{L}$ .
$L_2 = L_1\dfrac{f_1}{f_2} = 90 \times \dfrac{120}{180} = 60\text{ cm}$ .
So the answer is (c) $60\text{ cm}$ .

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