Velocity and Acceleration in S.H.M.

The velocity and acceleration of a particle executing S.H.M. are found directly by differentiating the displacement $x(t) = A\cos(\omega t + \phi)$ with respect to time -- once to get velocity, and a second time to get acceleration:\n\n$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi)$\n\n$a(t) = \frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t + \phi) = -\omega^2\,x(t)$\n\nThe last, compact form of the acceleration, $a=-\omega^2x$, restates the DEFINING property of S.H.M. directly in the language of kinematics: at every instant, the acceleration is exactly proportional to the displacement, and always directed opposite to it -- back towards the mean position -- which is exactly what a restoring force $F=-kx$ produces through Newton's second law, $F = ma$.\n\nFrom these expressions, the velocity has its greatest magnitude,\n\n$v_{\max} = A\omega$\n\nexactly AT the mean position ($x=0$, where the sine factor in $v(t)$ reaches its maximum magnitude of $1$), and falls all the way to zero at the two extreme positions ($x=\pm A$), where the particle momentarily stops before reversing direction. A useful relation connecting displacement and speed directly, without reference to time at all, is obtained by eliminating $t$ between $x(t)$ and $v(t)$ using the identity $\sin^2\theta+\cos^2\theta=1$:\n\n$v = \omega\sqrt{A^2 - x^2}$\n\nThe acceleration behaves in exactly the opposite way to velocity: it is zero exactly at the mean position (where the restoring force itself is zero, since $F=-kx=0$ when $x=0$), and reaches its greatest magnitude,\n\n$a_{\max} = A\omega^2$\n\nat the two extreme positions $x=\pm A$, where the displacement -- and hence the restoring force -- is largest.\n\nBecause $v(t)$ involves a sine function while $x(t)$ involves a cosine function of the very same underlying phase, velocity is said to LEAD displacement by a quarter cycle (a phase difference of $\pi/2$): the velocity curve reaches each of its own landmark values (zero, maximum, zero, minimu …