# A particle performs simple harmonic motion with amplitude A.  Its speed is trebled at the instant that it is at a distanc from equilibrium position.  The new amplitude of the motion is : Option 1) Option 2) Option 3) Option 4)

S Sabhrant Ambastha

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)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

Exams
Articles
Questions