A sound of frequency 480 Hz is emitted from the stringed instrument. The velocity of sound in air is 320 m/s. After completing 180 vibrations, the distance covered by a wave is (a) 60 m (b) 90 m (c) 120 m (d) 180 m

Answer: (c). In each vibration the wave advances one wavelength, so distance = Nλ = Nv/n. λ = (320)/(480) = 2/3 m Distance = 180 × 2/3 = 120 m So the answer is (c) 120 m.

A sound of frequency 480 Hz is emitted from the stringed instrument… | MHT CET 2023 (11 May, Shift 2)

Q21MHT CET 2023 · 11 MAY · SHIFT 2Easy

Wave Basics & Properties of Sound

A sound of frequency $480\text{ Hz}$ is emitted from the stringed instrument. The velocity of sound in air is $320\text{ m/s}$ . After completing 180 vibrations, the distance covered by a wave is

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Answer: (c)

In each vibration the wave advances one wavelength, so distance $= N\lambda = N\frac{v}{n}$ .
$\lambda = \frac{320}{480} = \frac{2}{3}$ m
Distance $= 180 \times \frac{2}{3} = 120$ m
So the answer is (c) 120 m.

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