Question

A particle is executing simple harmonic motion of amplitude $A$. At a distance $x$ from the centre, particle moving towards the extreme position received a blow in the direction of motion which instantaneously doubles the velocity. Its new amplitude will be

• A. $A$
• B. $\sqrt{A^2-x^2}$
• C. $\sqrt{2A^2-3x^2}$
• D. $\sqrt{4A^2-3x^2}$

Answer 1

The correct option is D $\sqrt{4A^2-3x^2}$

Velocity of particle in SHM is given as $v=\omega \sqrt{A^2-x^2}-\left(1\right)$

Let new amplitude be $A_1$,

as the velocity is doubled at distance $x$ from the center, with angular frequency $\omega$ remaining constant.

So, $2v=\omega \sqrt{A_1^2-x^2}-\left(2\right)$

From $\left(1\right)and\left(2\right)$, $A_1=\sqrt{4A^2-3x^2}$