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

Options

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

Answer

C

Explanation

We know that ` omega sqrt((k)/( M - m)) and omega _(1) = sqrt(( k)/(M)) (A_(i) = A)`
Also, momentum is conserved just befor and just after the block of mass (m) is placed because there is no impulsive force . So,
`MA_(i) omega _(i) = (M + m) v.`
`v. = (M A_(f) omega _(f))/((M + m)) , v. = A_(f) omega _(f)`
`(MA omega_(i))/( (M + m)) = A_(f) sqrt((k)/( (M + m)))`
`rArr (Ma sqrt((k)/(M)))/(M + m) xx sqrt((M + m)/( k)) = A_(f)`
`rArr A_(f) = A sqrt((M)/((M + m)))`
Aliter : On putting m on M. Let velocity becomes B `(m + M) V = MV_(0)` (by momentum conservation)
Now, kinetic Energy ` = (1)/(2) (m + M) V^(2)`
`K. = (1)/(2) (m + M) (M^(2) V_(0)^(2))/((m + M) ^(2))`
`K. = (1)/(2) (M^(2) V_(0)^(2))/( (M + m))` [` :. (1)/(2) MV_(0) ^(2) = (1)/(2) kA^(2) and K. = (1)/(2) k (A.)^(2)` ]
Hence `A. = sqrt((M)/( (M + m))) A `