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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

1.05 Hz

C

1050 Hz

D

10.5Hz

Text Solution

Verified by Experts

The correct Answer is:
A
Given `(nv)/(2l)=315 and (n+1)v/(2l)=420`
`implies (n+1)/n=420/315 implies n=3`
Hence `3 timesv/(2l)=315 implies v/(2l)=105Hz`
The lowest resonant frequency is when n=1
therefore lowest resonant frequency =105 Hz
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