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A spring elongated by length 'L' when a ...

A spring elongated by length 'L' when a mass 'M' is suspended to it. Now a tiny mass 'm' is attached and then released, its time period of oscillation is -

A

`2pisqrt(((M+m)l)/(Mg))`

B

`2pisqrt((ml)/(Mg))`

C

`2pisqrt(L//g)`

D

`2pisqrt((Ml)/((m+M)g))`

Text Solution

Verified by Experts

The correct Answer is:
A
`because T=2pisqrt(M/K) " " thereforeMg=Kl`
therefore `T = 2pisqrt(((M+m)l)/(Mg))`
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