Question

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

Answer 1

The Correct Option is A

The displacement equation of particle executing SHM is $x=a \cos (\omega t+\phi)$ ...(i) Velocity, $v=\frac{d x}{d t}=-a \omega \sin (\omega t+\phi)$ ...(ii)

$A=\frac{d v}{d t}=-a \omega^{2} \cos (\omega t+\phi)$ Fig. (i) is a plot of E (i) with $\phi=0$. Fig. (ii) shows E (ii) also with $\phi=0$. Fig. (iii) is a plot of E (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 $90^{\circ}(0.5 \pi)$ out of phase with the displacement and the acceleration is $90^{\circ}$ $(0.5 \pi)$ out of phase with the velocity but $180^{\circ}(\pi)$ out of phase with displacement.