A block with mass (M) is connected by a massless spring with stiffness constant (k) to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position (x_0). Consider two cases : (i) when the block is at (x_0) , and (ii) when the block is at `x = x_0 + A`. In both the cases, a particle with mass m(lt M) is softly placed on the block after which they strick to each other. Which of the following statement (s) is (are) true about the motion after the mass (m) is placed on the mass (M) ?

Options

A. The amplitude of oscillation in the first case changes by a factor of `sqrt((M)/(m+M))`, whereas in the second case it remains unchanged .
B. The final time period of oscillation in both the cases is same.
C. The total energy decreases in both the cases.
D. The instantaneous speed at `x_(0)` of the combined masses decreases in both the cases.

Answer

A, B, D

Explanation

When a block alonge oscillates on a horizontal frictionless surface, then its angular frequency of oscillation is, `omega_(0)=sqrt((k)/(M))`
After a mass `m` is placed on block, the angular frequency of oscillation , `omega=sqrt((k)/(m+M))`
Let `upsilon^(')` be the velocity of the blocks `(M+m)`, then using law of conservation of linear momentum,
we have
`Mupsilon_(0)=(M+m)upsilon^(')` or `Momega_(0)A=(M+m)omegaA^(')`
or `(A^('))/(A)=(M)/((M+m))xx(omega_(0))/(omega)`
`=(M)/((M+m))xx(sqrt(k//M))/(sqrt(k//(M+m)))=sqrt((M)/(M+m))`
At extreme position, the velocity of block,
`upsilon_(M)=0` ltbr. When mass `m` is placed on block os mass `M` at extreme position,
velocity of block and mass `m` is , `upsilon_(M+m)=0` ltbr. Therefore, the extreme position and mean position remain unchanged. Hence, `A^(')=A.`
Final time period of oscillation.
`T^(')=(2pi)/(omega)=2pisqrt((M+m)/(k))`.
It is same in both the cases.
Total energy decreases in 1st case, but not in 2nd case.
Instantaneous velocity at `x_(0), i.e.,`
Velocity at mean position `=omega A^(')`.. It decrease ini both the cases.