A mass M is attached to a horizontal spring of force constant k fixed one side to a rigid support as shown in figure. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass M is in equilibrium position, another mass m is gently placed on it. When will be the new amplitude of ocillation?

Options

A. `Asqrt((M)/(M-m))`
B. `Asqrt((M-m)/(M))`
C. `Asqrt((M)/(M+m))`
D. `Asqrt((M+m)/(M))`

Answer

C

Explanation

`T=2pisqrt((M)/(k))`
When the mass m is placed on mass M, the system of mass attached to the spring is `(M+m)`. New time period of oscillation,
`T.=2pisqrt((M+m)/(k))`
Let v be the velocity of the mass M while passing through the mean position and v. be the velocity of the mass `(M+m)` while passing through the mean position. Then
According to law of conservation of linear momentum
`Mv=(M+m)v.`
At mean position, `v=Aomegaandv.=A.omega.`
`therefore MAomega=(M+m)A.omega.`
or `A.=((M)/(M+m))(omega)/(omega.)A=(M)/(M+m)xx(T.)/(T)xxA`
`=((M)/(M+m))xxsqrt((M+m)/(M))xxA=Asqrt((M)/(M+m))`