Question

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 ${x_o}$; Consider two cases (i) when the block is at ${x_o}$; and (ii) when the block is $x = {x_o} + A$. In both cases, a particle with mass m(A. The amplitude of oscillation in the first case changes by a factor of $\sqrt {\dfrac{M}{{m + M}}} $ whereas in the second case it remains unchanged
B. The final time period of oscillation in both cases is same
C. The total energy decreases in both the cases
D. The instantaneous speed at ${x_o}$ of the combined masses decreases in both the cases

Answer 1

Hint:spring is connected with the rigid wall it means one of the side of the spring is fixed now you need to apply momentum conservation to find the velocity after the particle with mass m is gently placed on the block, then apply energy conservation to find the amplitude and then apply the formula of time period to compare the time periods.

Complete step by step answer:
Let's take case(i):
When the block is at ${x_o}$ and a particle is gently placed in it.
At equilibrium position velocity is maximum
Ie; ${u_1} = \sqrt {\dfrac{k}{M}} A$
After a particle is gently placed in it
Velocity of the whole system will be
$(M + m){v_1} = M{u_1}\\
\Rightarrow {v_1} = \dfrac{M}{{m + M}}\sqrt {\dfrac{k}{M}} A$
Because momentum will be conserved,
Now new amplitude $A' = \dfrac{{{v_1}}}{{\sqrt {\dfrac{k}{{m + M}}} }}$
$ \Rightarrow A' = \sqrt {\dfrac{M}{{M + m}}} A$
Time period depends upon total mass and stiffness constant only so it will remain the same in both cases as total mass is the same.

Now let's take case(ii):
Here particle is placed when the block is at extreme position where it has no velocity so there will be no change in velocity hence amplitude will remain same .And since amplitude is same so total energy of the system will remain same in this case .Now $\omega $ is changing in second case as external mass is added.
So new velocity at ${x_o}$ will be ${v_2} = \sqrt {\dfrac{k}{{M + m}}} A$ which is smaller than the original velocity

Hence Options-ABD are correct.

Note: In the second case amplitude is same and total mass of the system remains same but total energy will remain same because spring constant is same and the extra mass will result in reduction in the magnitude of velocity at every point on the path of the object.