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

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$\frac{(m_{1} + m_{2})^{2} g^{2}}{k}$

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Solution

The correct option is C

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

$m_{2} g - N = m_{2} a$ $w h e r e a = ω^{2} A$

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

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

$A = \frac{g}{ω^{2}}$ where $ω$ = angular frequency of oscillation of $(m_{1} + m_{2})$ and the spring.

Putting $ω^{2} = \frac{k}{m_{1} + m_{2}}$ , we obtain

$A = \frac{(m_{1} + m_{2}) g}{k}$

$∴ U_{m a x} = \frac{1}{2} k A^{2} = \frac{(m_{1} + m_{2})^{2} g^{2}}{2 k}$

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Q. A spring is loaded with two blocks

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m_{2}

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is rigidly fixed with the spring and

m_{2}

is just kept on the block

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as shown in the figure. The maximum energy of oscillation that is possible for the system having the block

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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