question_answer

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:                                                                                                                           [AIPMT (S) 2007]

A) \[0.5\,\,\pi \]

B)                   \[\pi \]

C)  \[0.707\,\,\pi \]

D)                   zero

Correct Answer: B

Solution :

The    displacement    equation    of   particle executing SHM is

\[x=a\cos (\omega t+\phi )\]                    ...(i)

velocity, \[v=\frac{dx}{dt}=-a\,\omega \sin \,(\omega t+\phi )\]                  (ii)

and acceleration,

\[A=\frac{dv}{dt}=-a{{\omega }^{2}}\,\cos \,(\omega t+\phi )\]        (iii)

Fig. (i) is a plot of Eq. (i) with \[\phi =0\]. Fig. (ii) shows Eq. (ii) also with \[\phi =0\]. 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 of A is shifted (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 \[{{90}^{o}}\,(0.5\,\pi )\] outo of phase with the velocity but \[{{180}^{o}}\,(\pi )\] out of phase with displacement.