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A string is stretched betweeb fixed poin...

A string is stretched betweeb fixed points separated by `75.0 cm`. It 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 strings is

A

`10.5 Hz`

B

`105 Hz`

C

`1.05 Hz`

D

`1050 Hz`

Text Solution

Verified by Experts

The correct Answer is:
2
`(n + 1) (v)/(2l) = 420 …..(1)`
`(nv)/(2l) = 315 ….(2)`
`(1) - (2) (V)/(mu) = 105 Hz , f_(min) = = 105 Hz`
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