Question

A spring is loaded with two blocks $m_1$ and $m_2$ where $m_1$ is rigidly fixed with the spring and $m_2$ is just kept on the block $m_1$ as shown in the figure. The maximum energy of oscillation that is possible for the system having the block $m_2$ in contact with $m_1$ is

• A. $\frac{m_1^2{g}^2}{2k}$
• B. $\frac{m_2^2{g}^2}{2k}$
• C. $\frac{(m_1+m_2{)}^2{g}^2}{2k}$
• D. $\frac{(m_1+m_2{)}^2{g}^2}{k}$

Answer 1

The correct option is C

$\frac{(m_1+m_2{)}^2{g}^2}{2k}$

Let the required amplitude be A. For this amplitude of oscillation the normal contact force between the blocks can be given as

$m_{2}g-N=m_{2}a$ $wherea={\omega }^{2}A$

$\Rightarrow m_{2}g-N=m_2{\omega }^{2}A$

Putting N = 0 for just losing contact for maximum amplitude, we obtain

$A=\frac{g}{{\omega }^2}$ where $\omega$ = angular frequency of oscillation of $(m_1+m_2)$ and the spring.

Putting ${\omega }^2=\frac{k}{m_1+m_2}$, we obtain

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

$\therefore U_{max}=\frac{1}{2}k{A}^2=\frac{(m_1+m_2{)}^2{g}^2}{2k}$