Simple Harmonic Motion: Common Mistakes and Fixes (1)

Question

A particle executes SHM with amplitude $A = 10\text{ cm}$ and time period $T = 2\text{ s}$. Find the speed of the particle when its displacement is $x = 5\text{ cm}$.

Solution — Step by Step

Why This Works

In SHM, total energy $\frac{1}{2}m\omega^2 A^2$ stays constant. At any instant, it splits between kinetic energy $\frac{1}{2}mv^2$ and potential energy $\frac{1}{2}m\omega^2 x^2$. Solving for $v$ gives the formula. At $x = 0$ (equilibrium), all energy is kinetic, so $v$ is maximum: $v_{\max} = A\omega$. At $x = A$, all energy is potential, so $v = 0$.

This is one of the most reliable tricks in oscillations: any “speed at displacement $x$” question collapses to $v = \omega\sqrt{A^2 - x^2}$.

Alternative Method

Differentiate $x = A\sin(\omega t)$:

$v = A\omega\cos(\omega t)$

When $x = A\sin(\omega t) = 5$, $\sin(\omega t) = 0.5$, so $\cos(\omega t) = \sqrt{3}/2$. Then $v = 10 \times \pi \times \sqrt{3}/2 = 5\pi\sqrt{3}\text{ cm/s}$. Same answer.

Common Mistake

Confusing time period $T$ with frequency $f$. The relation is $T = 1/f$ and $\omega = 2\pi f = 2\pi/T$. Plugging $f$ where $\omega$ belongs misses the $2\pi$ and gives a wildly wrong answer.