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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

1.05 Hz

B

1050 Hz

C

10.5 Hz

D

105 Hz

Text Solution

Verified by Experts

The correct Answer is:
D
`(nv)/(2l)=|315,((n+1)v)/(2l)=420`
`implies (n+1)/(n)=(420)/(315)impliesn=3`
Lowest resonant frequency is corresponds to n = 1
`therefore (v)/(2l)=(315)/(n)implies(v)/(2l)=(315)/(3)implies (v)/(2l)=105Hz`
Lowest resonant frequency = 105 Hz.
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