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Question

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is

A
105 Hz
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B
155 Hz
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C
205 Hz
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D
10.5 Hz
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Solution

The correct option is A 105 Hz
For string fixed at both ends,
f=nv2L

Length of the string L=75 cm

Since the given resonance frequencies are consecutive, we can write

xv2L=315
(x+1)v2L=420
for some integer x>1

Fundamental frequency f0=v2L=(x+1)v2Lxv2L=420315=105 Hz

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