Question

A mass M, attached to a horizontal spring executes S.H.M with amplitude ${{A}_{1}}$. When the mass M passes through its mean position then a smaller m is placed and both of them move together with amplitude ${{A}_{2}}$. The ratio of $\dfrac{{{A}_{1}}}{{{A}_{2}}}$ is :
A. ${{\left( \dfrac{M}{M+m} \right)}^{1/2}}$
B. ${{\left( \dfrac{M+m}{M} \right)}^{1/2}}$
C. $\dfrac{M}{M+m}$
D. $\dfrac{M+m}{M}$

Answer 1

Hint:To solve the given question we must two different concepts. One is conservation of momentum when the net force on the system is zero and the other is speed of the mass at the mean position under SHM. We must also express the angular frequency of the motion.

Formula used:
$v=A\omega $
$P=Mv$
$\omega =\sqrt{\dfrac{k}{m}}$

Complete step by step answer:
Let us use some properties of a spring block system under S.H.M (Simple harmonic motion). At the mean position (i.e. where the net force on the block is zero), the speed of the block is equal to $v=A\omega $,
where A is the amplitude of the S.H.M and $\omega $ is the angular frequency of the motion.

Let the speed of the single mass M, with amplitude ${{A}_{1}}$ and angular frequency ${{\omega }_{1}}$ be ${{v}_{1}}={{A}_{1}}{{\omega }_{1}}$.
$\Rightarrow {{A}_{1}}=\dfrac{{{v}_{1}}}{{{\omega }_{1}}}$ ….. (i)
Let the speed of system of the two masses M and m, with amplitude ${{A}_{2}}$ and angular frequency ${{\omega }_{2}}$ be ${{v}_{2}}={{A}_{2}}{{\omega }_{2}}$.
$\Rightarrow {{A}_{2}}=\dfrac{{{v}_{2}}}{{{\omega }_{2}}}$ ….. (ii).
Divide (i) by (ii)
$\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{\dfrac{{{v}_{1}}}{{{\omega }_{1}}}}{\dfrac{{{v}_{2}}}{{{\omega }_{2}}}}=\dfrac{{{v}_{1}}{{\omega }_{2}}}{{{v}_{2}}{{\omega }_{1}}}$ …. (iii).

Also, at the means position the net force on the mass M is zero and the mass m is also at rest. Therefore, this means that the net force for the system is zero at the mean position. Hence, the momentum of the system will be conserved when two masses collide with each.
Here, the mass M is moving with speed ${{v}_{1}}$ at the mean position. With this speed, it collides with the mass m and then both move together with speed ${{v}_{2}}$ in the same direction of ${{v}_{1}}$.

Therefore, the momentum of the system just before collision is ${{P}_{i}}=M{{v}_{1}}$.
The momentum of the system just after the collision is ${{P}_{f}}=(M+m){{v}_{2}}$.
But ${{P}_{i}}={{P}_{f}}$
$\Rightarrow M{{v}_{1}}=(M+m){{v}_{2}}$
$\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{(M+m)}{M}$
Substitute the ratio in (iii).
$\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\left( \dfrac{(M+m)}{M} \right)\dfrac{{{\omega }_{2}}}{{{\omega }_{1}}}$ …. (iv)

The angular frequency of a spring mass system is given as $\omega =\sqrt{\dfrac{k}{m}}$,
where k is the spring constant of the spring and m is the total mass attached to the spring.
Since in both the cases the spring remains the same, the value of k is the same in both.
In the first, the mass is M and in the second case the mass is $(M+m)$ .
This means that ${{\omega }_{1}}=\sqrt{\dfrac{k}{M}}$ …. (v)
and ${{\omega }_{2}}=\sqrt{\dfrac{k}{M+m}}$ …. (vi)
Now, divide (vi) and (v).
 $\Rightarrow \dfrac{{{\omega }_{2}}}{{{\omega }_{1}}}=\dfrac{\sqrt{\dfrac{k}{M+m}}}{\sqrt{\dfrac{k}{M}}}$
$\Rightarrow \dfrac{{{\omega }_{2}}}{{{\omega }_{1}}}=\sqrt{\dfrac{M}{M+m}}$
Substitute this value in (iv).
$\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\left( \dfrac{(M+m)}{M} \right)\sqrt{\dfrac{M}{M+m}}$
$\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\sqrt{\dfrac{M+m}{M}}\\
\therefore \dfrac{{{A}_{1}}}{{{A}_{2}}}={{\left( \dfrac{M+m}{M} \right)}^{\dfrac{1}{2}}}$.

Hence, the correct option is B.

Note: The formula $v=A\omega $ gives us the speed of the mass only at the mean position and this is the maximum value of the speed of the mass. Its speed varies in the sinusoidal function of time. However, a simple harmonic motion as a constant angular frequency.