Question

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 an equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be:

• A. $A\sqrt{\frac{M}{M+m}}$
• B. $A\sqrt{\frac{M}{M-m}}$
• C. $A\sqrt{\frac{M-m}{M}}$
• D. $A\sqrt{\frac{M+m}{M}}$

Answer 1

The correct option is A

$A\sqrt{\frac{M}{M+m}}$

Step 1: Given information

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 $T$, amplitude $A$, and another mass $m$ is gently fixed upon it.

Step 2: Draw the required diagram

Draw the diagram before placing the mass $m$, and after placing mass $m$.

Step 3: Calculate the new amplitude of oscillation

The general formula of angular frequency in terms of propagation constant and mass is $\omega =\sqrt{\frac{k}{m}}$

Thus, the angular frequency in the initial state is,

${\omega }_i=\sqrt{\frac{k}{M}}$

Similarly, the angular frequency in the final state is,

${\omega }_f=\sqrt{\frac{k}{M+m}}$

Additionally, because there is no impulsive force, momentum is conserved immediately before and immediately after the block of mass $\left(m\right)$. So,

$\begin{array}{l}\Rightarrow M{A}_i{\omega }_i=\left(M+m\right)v\text{'}\\ \Rightarrow v\text{'}=\frac{M{A}_i{\omega }_i}{\left(M+m\right)}\\ \Rightarrow v\text{'}=A_f{\omega }_f\end{array}$

Therefore,

$\begin{array}{l}\Rightarrow \frac{MA{\omega }_i}{\left(M+m\right)}=A_f\sqrt{\frac{K}{\left(M+m\right)}}\\ \Rightarrow \frac{MA\sqrt{\frac{K}{M}}}{\left(M+m\right)}\times \sqrt{\frac{M+m}{K}}=A_f\\ \Rightarrow A_f=A\sqrt{\frac{M}{M+m}}\end{array}$

Hence, the correct answer is option (A).