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A string is stretched between fixed poin...

A string is stretched between fixed points separated by `75.0cm`. 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

105 Hz

B

155 Hz

C

205 Hz

D

10.5 Hz

Text Solution

Verified by Experts

The correct Answer is:
A
`315:420=3:4`
315 Hz is a `3^(rd)` harmonic of wire and 420 Hz
is a `4^(th)` harmonic.
if 315 is a `3^(rd)` harmonic then `3n_("fundamental")=315`
`n_("fundamental")=(315)/(3)=105Hz`
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