Let angular frequency of SHM is ω and when is at distance x from mean position, its velocity will be given as v=ω√A2−x2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) When its velocity becomes double, let new amplitude is A′. So, ‌‌2v=ω√A′2−x2 Putting the value of v from equation (i) ‌2ω√A2−x2=ω√A′2−x2 ‌2√A2−x2=√A′2−x2 ⇒4(A2−x2)=A′2−x2 ⇒A′2=4A2−3x2⇒A′=√4A2−3x2
