A point oscillates according to x = a cos omega t. Find

A point oscillates according to $x = a\cos\omega t$. Find the velocity and acceleration at $x = \pm a$ and at $x = 0$, and show that $\dfrac{d^{2}x}{dt^{2}} = -\omega^{2}x$.

Step 1: Velocity and acceleration

\[ v = \frac{dx}{dt} = -a\omega\sin\omega t, \qquad w = \frac{d^{2}x}{dt^{2}} = -a\omega^{2}\cos\omega t. \]

Step 2: At the extreme positions $x = \pm a$

There $\cos\omega t = \pm1$, which forces $\sin\omega t = 0$: \[v = 0, \qquad w = -a\omega^{2}\bigl(\pm1\bigr) = \mp a\omega^{2}.\]

The point is momentarily at rest, with acceleration of maximum magnitude pointing back towards the centre.

Step 3: At the centre $x = 0$

There $\cos\omega t = 0$, so $\sin\omega t = \pm1$: $v = \mp a\omega, \qquad w = 0$.

Speed is greatest exactly where acceleration vanishes — the two extremes never coincide.

Step 4: The equation of motion

Since $a\cos\omega t$ is $x$ itself, \[w = -a\omega^{2}\cos\omega t = -\omega^{2}\bigl(a\cos\omega t\bigr) = -\omega^{2}x.\]

Step 5: Verification

\[v^{2}+\omega^{2}x^{2} = a^{2}\omega^{2}\sin^{2}\omega t+a^{2}\omega^{2}\cos^{2}\omega t = a^{2}\omega^{2} \ \text{ (constant)} \;\;\checkmark\]

\[\text{Setting } x = 0 \text{ gives } |v| = a\omega \;\;\checkmark; \quad \text{setting } v = 0 \text{ gives } |x| = a \;\;\checkmark\]

\[\text{Period } T = \frac{2\pi}{\omega}; \text{ the four listed states recur once per period} \;\;\checkmark\]