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A mass $M$ attached to a horizontal spring, executes $SHM$ with a amplitude $A_{1}$ When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_{2}$ . The ratio of $\left ( \frac{A_{1}}{A_{2}} \right )$ is

Option 1)
$\frac{M}{M+m }$

Option 2)
$\frac{M+m}{M }$

Option 3)
$\left ( \frac{M}{M+m } \right )^{1/2}$

Option 4)
$\left ( \frac{M+m}{M } \right )^{1/2}$

Answers (1)

best_answer

As we learnt in

Time period of oscillation for spring mass system -

$T= 2\pi \sqrt{\frac{m}{K}}$

- wherein

m = mass of block

K = spring constant

From conservation of linear momentum

$\overrightarrow{p_{i}}=\overrightarrow{p_{f}}\ \; \Rightarrows\ \; M_{y}=(M+m)v_{1}$

$\frac{v}{v_{1}}=\frac{M+m}{M}\ \; \Rightarrow\ \; \frac{A\omega}{A_{1}\omega_{1}}=\frac{v}{v_{1}}$

$\frac{A_{1}}{A}=\frac{\omega \times v_{1}}{\omega_{1}\times v}$ $\left[\because\ \; \omega=\sqrt{\frac{k}{m}} \right ]$

$\frac{\omega}{\omega_{1}}=\sqrt{\frac{\frac{k}{M}}{\frac{k}{M+m}}}=\sqrt{\frac{M+m}{m}}$

$\therefore\ \; \frac{A_{1}}{A}=\sqrt{\frac{M+m}{m}}\times\frac{M}{M+m}$

$\Rightarrow\ \; \frac{A_{1}}{A}=\sqrt{\frac{M}{M+m}}$

Correct option is 4.

Option 1)

$\frac{M}{M+m }$

This is an incorrect option.

Option 2)

$\frac{M+m}{M }$

This is an incorrect option.

Option 3)

$\left ( \frac{M}{M+m } \right )^{1/2}$

This is an incorrect option.

Option 4)

$\left ( \frac{M+m}{M } \right )^{1/2}$

This is the correct option.

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