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Question: A mass M is suspended from a light spring. An additional mass m added displaces the spring further b...

A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period

A

T=2π(mg/x(M+m))T = 2 \pi \sqrt { ( m g / x ( M + m ) ) }

B

T=2π((M+m)x/mg)T = 2 \pi \sqrt { ( ( M + m ) x / m g ) }

C

T=(π/2)(mg/x(M+m))T = ( \pi / 2 ) \sqrt { ( m g / x ( M + m ) ) }

D

T=2π((M+m)/mgx)T = 2 \pi \sqrt { ( ( M + m ) / m g x ) }

Answer

T=2π((M+m)x/mg)T = 2 \pi \sqrt { ( ( M + m ) x / m g ) }

Explanation

Solution

As mg produces extension x, hence k=mgxk = \frac { m g } { x }

∴ T=2π(M+m)k=2π(M+m)xmgT = 2 \pi \sqrt { \frac { ( M + m ) } { k } } = 2 \pi \sqrt { \frac { ( M + m ) x } { m g } }