A body executing SHM with time period T and amplitude A . When it is at +A/2 from equilibrium position moving towards right . It receives an impulse

Question

which doubles its  velocity without changing direction .

Find (a) velocity after the blow

(B) amplitude after the amplitude

(C) time taken by the block to reach the right extreme position

Answer 1

Velocity of particle in SHM is given as v=ωA2−x2​ −(1)

Let new amplitude be A1​,

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

So, 2v=ωA12​−x2​ −(2)

From (1) and (2), A1​=4A2−3x2​

Comments

I didn't get how u have done .........all the characters mentioned here are irrelevant