Explanation

(a) 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)
Acceleration,
[image:14_image50.emf]
$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 $(1/4T).$ Similarly, the acceleration curve of $A$ is shifted (to the left) by $1/4T$ 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