JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Waves & Sound

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A string 2.0 m long and fixed at its end is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is:

(A) $320 m / s, 120 Hz$
(B) $180 m / s, 80 Hz$
(C) $180 m / s, 120 Hz$
(D) $320 m / s, 80 Hz$ ✓ Correct

Answer: (D)

Solution

In its third harmonic a string fixed at both ends satisfies

f_{3} = \frac{3 v}{2 L}

. With

f_{3} = 240

Hz and

L = 2.0

240 = \frac{3 v}{4}

, so

v = 320

m/s. The fundamental is

f_{1} = \frac{v}{2 L} = \frac{320}{4} = 80

Hz, one-third of the driving frequency as expected.

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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