A particle performs simple harmonic motion with amplitude A.  Its speed is trebled at the instant that it is at a distanc

\frac{2A}{3} from equilibrium position.  The new amplitude of the motion is :

  • Option 1)

    \frac{A}{3}\sqrt{41}

  • Option 2)

    3A

  • Option 3)

    A\sqrt{3}

  • Option 4)

    \frac{7A}{3}

 

Answers (1)

As we learnt in

Relation of velocity and displacement -

v= w\sqrt{A^{2}-x^{2}}
 

- wherein

\rightarrow  x is displacement from mean position

\rightarrow  A is Amplitude.

 

We know that 

     V=\omega\sqrt{A^{2}-x^{2}},            x=\frac{2A}{3}

Initially - V=\omega \sqrt{A^{2}-\left(\frac{24}{3} \right )^{2}}

Finally - 3V=\omega \sqrt{A^{2}-\left(\frac{24}{3} \right )^{2}}                A'= Final amplitude

Now dividing we get \frac{3}{1}=\frac{\sqrt{A'^{12}-\left(\frac{2A}{3} \right )^{2}}}{\sqrt{A^{2}- \left(\frac{2A}{3} \right )^{2}}}\ \; \Rightarrow\ \; 9 \left(\frac{A^{2}-4A^{2}}{9} \right )=\frac{A'^{2}-4A^{2}}{9}

\therefore    After calculate we get

    A'=\frac{7A}{3}

Correct option is 4.

 

    

 


Option 1)

\frac{A}{3}\sqrt{41}

This is an incorrect option.

Option 2)

3A

This is an incorrect option.

Option 3)

A\sqrt{3}

This is an incorrect option.

Option 4)

\frac{7A}{3}

This is an incorrect option.

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