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  4. Two masses m1=1 kg and m2=0.5 kg are suspended together by a massless spring of spring constant 12.5 Nm -1. When masses are in equilibrium m1 is removed without disturbing the system. New amplitude of oscillation will be

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Q. Two masses $m_{1}=1\, kg$ and $m_{2}=0.5\, kg$ are suspended together by a massless spring of spring constant $12.5\, Nm ^{-1}$. When masses are in equilibrium $m_{1}$ is removed without disturbing the system. New amplitude of oscillation will be

Oscillations

Solution:

Points of equilibrium of the spring will be when no force acts on it.
or $k x=\left(m_{1}+m_{2}\right) g$
$x=\frac{\left(m_{1}+m_{2}\right) g}{k}$
The new equilibrium position which will be the mean position of S.H.M. will be simply $\frac{m_{2} g}{k}$
New amplitude will be maximum displacement from $\frac{m_{2} g}{k}$ which is :
$A=\frac{\left(m_{1}+m_{2}\right) g}{k}-\frac{m_{2} g}{k}$
or $A=\frac{m_{1} g}{k}$
or $A=\frac{1 \times 10}{12.5}$
or $A=\frac{4}{5} m$
$\therefore A=0.8 \,m$ or $80 \,cm$