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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
Two consecutive resonant frequencies for a string fixed at both ends will be
`(nv)/(2l)` and `((n+1)v)/(2l)`
`implies((n+1)v)/(2l)-(nv)/(2l)=420-315`
`implies(v)/(2l)=105Hz`. Which is the minimum resonant frequency.
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