Question

A particle is executing simple harmonic motion of amplitude $A$. At a distance $x$ from the centre, a particle moving towards the extreme position receives 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 $\sqrt{4A^2-3x^2}$

we know the formula of rotational and translation motion,

$\frac{1}{2}K{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}K{x}^2$

after some time when its time get doubles the v' = 2v

$\frac{1}{2}K{A}^{\prime 2}=\frac{1}{2}m{v}^{\prime 2}+\frac{1}{2}K{x}^2$

$\frac{1}{2}K{A}^{\prime 2}=\frac{1}{2}m(2v{)}^2+\frac{1}{2}K{x}^2$

$A^{\prime }=\sqrt{4A^2-3x^2}$