question_answer

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

A) \[0.5\,\pi \]         

B) \[\pi \]

C) \[0.707\pi \]      

D)  zero

Correct Answer: A

Solution :

 The displacement equation of particle executing SHM is \[x=a\cos \,\omega t\] (i) Velocity, \[v=\frac{dx}{dt}=-a\omega \sin \omega t\] (ii) and its acceleration, \[a=\frac{dv}{dt}=-a{{\omega }^{2}}\cos \omega t\] (iii) and its acceleration, Fig. (i) shows a plot of Eq. (i). Fig. (ii) shows Eq. (ii) and Fig (iii) is a plot of Eq. (iii). It should be noted that in the figures the curve of v is shifted (to the left) from the curve of\[x\]by one-quarter period \[\left( \frac{1}{4}T \right).\] Similarly, the acceleration curve, stuffed (to the left) by\[\frac{1}{4}T\] relative to the velocity curve of v. This implies that velocity is \[{{90}^{o}}(0.5\pi )\]out of phase with the displacement and the acceleration is \[{{90}^{o}}(0.5\pi )\]out of phase with the velocity but \[{{180}^{o}}(\pi )\] out of phase with displacement.