Numerical 1: SHM — Find velocity and acceleration

Problem: A particle executes SHM with amplitude A = 10 cm and time period T = 2 s. Find: (a) maximum velocity, (b) maximum acceleration, (c) velocity at x = 6 cm.

Step 1 — Angular frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{2\text{ s}} = \pi \text{ rad/s}$

Step 2 — Convert amplitude to SI: $A = 10\text{ cm} = 0.10\text{ m}$

Step 3(a) — Maximum velocity (at x = 0): $v_{max} = A\omega = 0.10\text{ m} \times \pi\text{ rad/s} = 0.314\text{ m/s} \approx 31.4\text{ cm/s}$

Step 3(b) — Maximum acceleration (at x = ±A): $a_{max} = A\omega^2 = 0.10\text{ m} \times (\pi\text{ rad/s})^2 = 0.10 \times 9.87 = 0.987\text{ m/s}^2$

Step 3(c) — Velocity at x = 6 cm = 0.06 m: $v = \omega\sqrt{A^2 - x^2} = \pi\sqrt{(0.10)^2 - (0.06)^2}\text{ m}$ $= \pi\sqrt{0.0100 - 0.0036}\text{ m} = \pi\sqrt{0.0064}\text{ m}$ $= \pi \times 0.08\text{ m} = 0.251\text{ m/s} = 25.1\text{ cm/s}$