13.Oscillations
medium

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 .......... $cm$

A

$30$

B

$50$

C

$80$

D

$60$

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 $80 \,cm$

Standard 11
Physics

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