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 equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it then 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 $A\sqrt{\frac{M}{M+m}}$

Before placing the mass $m$, angular frequency,
${\omega }_i=\sqrt{\frac{k}{M}}$

After placing the mass $m$, angular frequency,
${\omega }_f=\sqrt{\frac{k}{M+m}}$

As there is no impulsive force, so, linear momentum will remain conserved.

$\therefore p_i=p_f$
$\Rightarrow M{\omega }_i{A}_i=(M+m){\omega }_f{A}_f$

$\Rightarrow M\times \sqrt{\frac{k}{M}}\times A=(M+m)\times \sqrt{\frac{k}{M+m}}\times A_f$

$\Rightarrow A_f=A\sqrt{\frac{M}{M+m}}$