Physics - SHM Question with Solution | TestHub

Let the spring constant be $k$. The velocity of the mass $m$ at the mean position is maximum and is given by $v_1=A_1{\omega }_1=A_1\sqrt{\frac{k}{m}}$.

When another mass $m$ is gently placed on it at the mean position, the net external force in the horizontal direction is zero. By conservation of linear momentum:

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

$v_2=\frac{v_1}{2}$

The new system has a total mass of $2m$, so its new angular frequency is ${\omega }_2=\sqrt{\frac{k}{2m}}$.

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

Substituting the values of $v_2$ and ${\omega }_2$:

$\frac{v_1}{2}=A_2\sqrt{\frac{k}{2m}}$

$\frac{1}{2}{A}_1\sqrt{\frac{k}{m}}=A_2\sqrt{\frac{k}{2m}}$

$\frac{A_1}{2}=\frac{A_2}{\sqrt{2}}$

$\frac{A_1}{A_2}=\frac{2}{\sqrt{2}}=\sqrt{2}$

Answer: $\sqrt{2}$