Question

A spring is loaded with two blocks $m_{1}and{m}_2$, where $m_1$ is just kept on the block $m_1$. The maximum energy of oscillation is possible for the system having the block $m_2$ in contact with $m_1$ is

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

Answer 1

The correct option is D $\frac{(m_1+m_2{)}^2{g}^2}{2k}$

Let x be the required amplitude of oscillation. For this amplitude of oscillation the normal contact force between the blocks can be given by
$R-m_{2}g=m_{2}a$
where, acceleration
$a=-{\omega }^{2}x$
$\therefore R-m_{2}g=-m_2{\omega }^{2}x$
For maximum amplitude of oscillation without loosing contact of block $m_2$ with block $m_1,R=0$
we get, $x=g/{\omega }^2$
where $\omega$ is the angular frequency of oscillation of $(m_1+m_2)$ and spring. Therefore
$\omega =\sqrt{\frac{k}{(m_1+m_2)}}$ or ${\omega }^2=\frac{k}{(m_1+m_2)}$
$\Rightarrow x=\frac{(m_1+m_2)g}{k}$
Hence, the maximum energy of oscillation
$U_{max}=\frac{1}{2}k{x}^2=\frac{(m_1+m_2{)}^2{g}^2}{2k}$