$A$ body describes simple harmonic motion with an amplitude of $5\; cm$ and a period of $0.2\; s$. Find the acceleration and velocity of the body when the displacement is $(a)\; 5\; cm$,$(b)\; 3\; cm$,and $(c)\; 0\; cm$.

(N/A) Given: Amplitude $A = 5\; cm = 0.05\; m$,Time period $T = 0.2\; s$. Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi\; rad/s$.
$(a)$ For displacement $x = 5\; cm = 0.05\; m$:
Acceleration $a = -\omega^2 x = -(10\pi)^2 \times 0.05 = -100\pi^2 \times 0.05 = -5\pi^2\; m/s^2$.
Velocity $v = \omega \sqrt{A^2 - x^2} = 10\pi \sqrt{(0.05)^2 - (0.05)^2} = 0\; m/s$.
$(b)$ For displacement $x = 3\; cm = 0.03\; m$:
Acceleration $a = -\omega^2 x = -(10\pi)^2 \times 0.03 = -100\pi^2 \times 0.03 = -3\pi^2\; m/s^2$.
Velocity $v = \omega \sqrt{A^2 - x^2} = 10\pi \sqrt{(0.05)^2 - (0.03)^2} = 10\pi \sqrt{0.0025 - 0.0009} = 10\pi \sqrt{0.0016} = 10\pi \times 0.04 = 0.4\pi\; m/s$.
$(c)$ For displacement $x = 0\; m$:
Acceleration $a = -\omega^2 x = 0\; m/s^2$.
Velocity $v = \omega \sqrt{A^2 - x^2} = 10\pi \sqrt{(0.05)^2 - 0} = 10\pi \times 0.05 = 0.5\pi\; m/s$.