A mass M attached to a horizontal spring, executes SHM with a amplitude A_{1} When the massM 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)

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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