- Home
- Standard 11
- Physics
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 .......... $cm$
$30$
$50$
$80$
$60$
Solution
(c)
Points of equilibrium of the spring will be when no force acts on it.
$k x=\left(m_1+m_2\right) g$
$x=\frac{\left(m_1+m_2\right) g}{k}$
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 $80 \,cm$