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A rod of mass 'M' and length '2L' is sus...

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations, if two masses each of 'm' are attached at distance` 'L//2' ` from its centre on both sides, it reduces the oscillation frequency by `20%`. The value of ratio `m//M` is close to :

A

`0.57`

B

`0.17`

C

`0.77`

D

`0.37`

Text Solution

Verified by Experts

The correct Answer is:
D

`f_(0)=(1)/(2pi)sqrt((c)/(I_(0)))`
`f_(0)=(1)/(2pi)sqrt((3c)/(ML^(2)))` .............(i)
`I.=(ML^(2))/(3)+2.(mL^(2))/(4)`
`f.=(1)/(2pi)sqrt((c)/(L^(2)((M)/(3)+(m)/(2))))=0.8f_(0)`............(ii)
From equation (i) and (ii)
`(m)/(M)=(9)/(24)=(3)/(8)=0.375`
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