JEE Main · 2006 · 30 Apr 2006 · Offline · Waves & Sound

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A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

(A)10.5 Hz
(B)105 Hz✓ Correct
(C)1.05 Hz
(D)1050 Hz

Answer: (B)

Solution

For a string fixed at both ends the resonant frequencies are whole-number multiples of the fundamental

f_{1}

. Two neighbouring resonances with nothing in between differ by exactly one fundamental, so

f_{1} = 420 - 315 = 105 Hz

. As a check,

315 = 3 f_{1}

and

420 = 4 f_{1}

, which are indeed consecutive. The lowest resonant frequency of the string is therefore

105 Hz

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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