Question

Pure masses $m_1$ and $m_2$ are suspended together by a mass less spring of constant K. given the mass are in equilibrium $m_2$ is removed without disturbing the system,the amplitude of oscillation will be

Answer 1

At equilibrium,

Force = Spring constant x extension = $kx$

Weight = $(m_1+m_2)g$

At equilibrium Spring force is equal to weight of masses

$k{x}_1=(m_1+m_2)g$

$x_1=\frac{(m_1+m_2)g}{k}$

If mass $m_2$ is removed, then at equilibrium extension of spring become.

$x_2=\frac{m_{1}g}{k}$

but, if $m_2$ cut out suddenly then, Amplitude of oscillation is the difference in extension

$x_1-x_2=\frac{(m_1+m_2)g}{k}-\frac{m_{1}g}{k}=\frac{m_{2}g}{k}$

Amplitude of oscillation $\frac{m_{2}g}{k}$