Question

Two masses $m_1=1\text{kg}$ and $m_2=0.5\text{kg}$ are suspended together by a massless spring of spring constant $12.5\text{N/m}$. When masses are in equilibrium, $m_1$ is removed without disturbing the system. New amplitude of oscillation will be

• A. $30\text{cm}$
• B. $50\text{cm}$
• C. $80\text{cm}$
• D. $60\text{cm}$

Answer 1

The correct option is C $80\text{cm}$

Given that,
Mass, $m_1=1\text{kg}$
Mass, $m_2=0.5\text{kg}$
Spring constant, $k=12.5\text{N/m}$
For the combination of $m_1$ and $m_2$,
$kx=(m_1+m_2)g$
$\Rightarrow x=\frac{(m_1+m_2)g}{k}$
$[$extension in spring$]$
After removal of mass $m_1$,
$k{x}^{\prime }=m_{2}g$
$\Rightarrow x^{\prime }=\frac{m_{2}g}{k}$

Thus, the amplitude of oscillation is,
$A=x-x^{\prime }$
$\Rightarrow A=\frac{(m_1+m_2)g}{k}-\frac{m_{2}g}{k}$
$\Rightarrow A=\frac{m_{1}g}{k}$
$\Rightarrow A=\frac{1\times 10}{12.5}=0.8\text{m}=80\text{cm}$