Simple Harmonic Motion: Step-by-Step Worked Examples (4)

Question

A particle executes SHM with amplitude $A = 5 \, \text{cm}$ and time period $T = 2 \, \text{s}$. Find its speed when it is at a displacement of $3 \, \text{cm}$ from the mean position.

Solution — Step by Step

Why This Works

The formula $v = \omega\sqrt{A^2 - x^2}$ encodes everything we need to know about SHM kinematics in one line. At the extremes ($x = \pm A$), $v = 0$. At the mean position ($x = 0$), $v$ is maximum at $\omega A$.

The 3-4-5 triangle showing up here is no coincidence — JEE/NEET examiners love amplitudes and displacements that form Pythagorean triples ($3, 4, 5$ or $5, 12, 13$).

Alternative Method — Differentiation

We could write $x = A \sin(\omega t + \phi)$, find $t$ when $x = 3$, then differentiate to get $v = A\omega \cos(\omega t + \phi)$. The trigonometric identity $\sin^2 + \cos^2 = 1$ gives the same result. Slower path, same destination.

For exams, memorise $v = \omega\sqrt{A^2 - x^2}$ and skip the trigonometry.

Common Mistake

Students often confuse $\omega$ (angular frequency) with $f$ (frequency) and write $v = f\sqrt{A^2 - x^2}$. This is off by a factor of $2\pi$ — fatal in any timed paper.

NEET adds 1-2 SHM problems every year — pure scoring topic if formulas are at our fingertips.