Question

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

Answer 1

A $=$ 5 cm $=$ 0.05 m
T $=$ 0.2 s
$\omega =2\pi /T=2\pi /0.2=10\pi rad/s$

When displacement is y, then acceleration, $a=-{\omega }^{2}y$

Velocity, $V=\omega \sqrt{r^2-y^2}$

Case (a) When $y=5cm=0.05m$

$a=-(10\pi {)}^2\times 0.05=-5{\pi }^{2}m/s^2$

$V=10\pi \times \sqrt{(0.05{)}^2-(0.05{)}^2}=0$

Case (b) When $y=3cm=0.03m$

$a=-(10\pi {)}^2\times 0.03=-3{\pi }^{2}m/s^2$

$V=10\pi \times \sqrt{(0.05{)}^2-(0.03{)}^2}=10\pi \times 0.04=0.4\pi m/s$

Case (c) When $y=0$

$a=-(10\pi {)}^2\times 0=0$

$V=10\pi \times \sqrt{(0.05{)}^2-0^2}=10\pi \times 0.05=0.5\pi m/s$