Oscillations and SHM: Exam-Pattern Drill (4)

Question

A particle executes SHM with amplitude $A = 0.1$ m and time period $T = 2$ s. Find (a) its maximum speed, (b) its speed when displacement is $0.05$ m, and (c) its acceleration at the extreme position.

Solution — Step by Step

Final answers: $v_{\max} \approx 0.314$ m/s, $v \approx 0.272$ m/s, $a_{\max} \approx 0.987$ m/s$^2$.

Why This Works

SHM is fully described by $\omega$ and $A$. Once these are pinned down, every other quantity — speed, acceleration, energy — follows from standard relations. The energy conservation equation $\frac{1}{2}m\omega^2 A^2 = \frac{1}{2}m v^2 + \frac{1}{2}m\omega^2 x^2$ is what gives us $v(x)$.

This is a frequent NEET pattern: give $A$ and $T$, ask for $v$ at some $x$. The answer is always $\omega \sqrt{A^2 - x^2}$.

Alternative Method

Energy approach. Total energy $E = \frac{1}{2}m\omega^2 A^2$. Kinetic energy at $x$ is $E - \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$. Setting this equal to $\frac{1}{2}mv^2$ recovers the same $v$.

Common Mistake

Mixing up $v_{\max}$ and $a_{\max}$ locations. $v_{\max}$ is at the mean position (where $x = 0$); $a_{\max}$ is at the extremes (where $|x| = A$). Students often write $v_{\max}$ at the extreme — it is actually zero there.