Question

A particle describes simple ahrmonic motion with amplitude of 5 cm and period of 0.2 sec. Calculate the velocity and acceleration of the particle when the displacement is 0 cm

Answer 1

For simple harmonic motion

$x(t)=A\sin(\omega t)\to x(t)=0.05\cdot\sin(10\pi t)$

$v(t)=0.05\cdot(10\pi)\cdot\cos(10\pi t)=0.5\pi\cdot\cos(10\pi t)$

$a(t)=-0.5\pi\cdot10\pi\cdot\sin(10\pi t)=-5\pi^2\cdot\sin(10\pi t)$

If $x(0)=0$ then $t=0$ . So,

$v=0.5\cdot3.14=1.57\ (m/s)$ ;

$a=0$ .