Physics - SHM Question with Solution | TestHub

Velocity of mass $m_1$ at the mean position is given by $v=A\omega =A\sqrt{\frac{k}{m_1}}$.

When mass $m_2$ is placed on $m_1$, the horizontal momentum of the system is conserved.

$m_{1}v=(m_1+m_2)v_1$

$v_1=\frac{m_{1}v}{m_1+m_2}$

The new angular frequency of the system is ${\omega }_1=\sqrt{\frac{k}{m_1+m_2}}$.

The new amplitude $A_1$ is related to the new maximum velocity $v_1$ by $v_1=A_1{\omega }_1$.

Substituting the values of $v_1$ and ${\omega }_1$:

$A_1\sqrt{\frac{k}{m_1+m_2}}=\frac{m_1}{m_1+m_2}A\sqrt{\frac{k}{m_1}}$

$A_1=A\frac{m_1}{m_1+m_2}\sqrt{\frac{m_1+m_2}{m_1}}$

$A_1=A\sqrt{\frac{m_1}{m_1+m_2}}$

Rearranging to find the ratio $\frac{A}{A_1}$:

$\frac{A}{A_1}={\left[\frac{m_1+m_2}{m_1}\right]}^{\frac{1}{2}}$

Answer: ${\left[\frac{m_1+m_2}{m_1}\right]}^{\frac{1}{2}}$