Question

The phase difference between the displacement and acceleration of particle executing S.H.M. radian is:

• A. $\pi /4$
• B. $\pi /2$
• C. $\pi$
• D. $2\pi$

Answer 1

Answer: (C)

$\pi$

$va\sin (\omega t+\varphi )$

Differentiating

$v=a\omega \cos (\omega t+\varphi )$

Differentiating again,

$a=-a{\omega }^2\sin (\omega +\varphi )$

Now taking minus sign

$a=a{\omega }^2\sin (\omega t+\varphi +\pi )$

So, phase difference between displacement and acceleration will be:-

$=\omega t+\varphi +\pi -\omega t+\varphi =\omega t+\varphi +\pi -\omega t-\varphi =\pi$