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For a string, tied with two fixed rigid supports, scparated by 75 cm has two consecutive harmonics at 315 Hz and 420 Hz. Find its minimum frequency.

A

1050 Hz

B

`10.5 Hz`

C

`105 Hz`

D

`1.05 Hz`

Text Solution

Verified by Experts

The correct Answer is:
C
We have, `f = (nv)/(2L ) = (n)/(2L) sqrt ((T)/(mu)) = 420 Hz ...(1)`
`f.= (n .v)/(2L) = (n.)/(2L) sqrt ((T)/(mu)) = 315 Hz " "...(2)`
`therefore (f)/(f..) = (n)/(n.) =(420)/(315) = 4/3`
When `n=4, n.=3,`
From equation (1),
`(4v)/(2L) = 420`
`therefore (v)/(2L) = 105 Hz`
`therefore f _(1) = f _(min) = 105 Hz " "(because f _(1) = (v)/( 2L))`
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