Physics - SHM Question with Solution | TestHub

Let the spring constant be $k$.

The maximum velocity of mass $M$ at the mean position is $v_1=A_1{\omega }_1=A_1\sqrt{\frac{k}{M}}$.

When mass $m$ is placed on $M$ at the mean position, the collision is perfectly inelastic. By conservation of linear momentum:

$M{v}_1=(M+m)v_2$

$v_2=\frac{M}{M+m}{v}_1$

The new angular frequency of the system is ${\omega }_2=\sqrt{\frac{k}{M+m}}$.

The new amplitude $A_2$ is related to the new maximum velocity by $v_2=A_2{\omega }_2$.

Substituting the values, we get:

$A_2\sqrt{\frac{k}{M+m}}=\frac{M}{M+m}{A}_1\sqrt{\frac{k}{M}}$

$A_2=A_1\left(\frac{M}{M+m}\right)\sqrt{\frac{M+m}{M}}$

$A_2=A_1\sqrt{\frac{M}{M+m}}$

$\frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}={\left(\frac{M+m}{M}\right)}^{\frac{1}{2}}$

Answer: ${\left(\frac{M+m}{M}\right)}^{\frac{1}{2}}$