Physics - SHM Question with Solution | TestHub

Take$x=A\sin \omega t$.

Then$v=A\omega \cos \omega t,a=-A{\omega }^2\sin \omega t$.

(A) acceleration is the inverted displacement graph - correct.

(B) velocity is a cosine, leading displacement by T/4 - correct.

$K.E.=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega t,P.E.=\frac{1}{2}k{A}^2{\sin }^2\omega t$

(C) KE is always positive with period T/2 - correct.

(D) the PE graph shown does not match${\sin }^2\omega t$- wrong.

Hence (A),(B),(C).