Question

A particle is executing simple harmonic motion. Its displacement is given by $x=5\sin \pi t$ where $x$ is in $\text{cm}$ and $t$ in seconds. How long will the particle take to move from the position of equilibrium to the position of maximum displacement ?

• A. $0.5\text{s}$
• B. $1.0\text{s}$
• C. $1.5\text{s}$
• D. $2.0\text{s}$

Answer 1

The correct option is A $0.5\text{s}$

Maximum displacement = amplitude $=5\text{cm}$
At time $t=0,x=0$ (equilibrium position). Hence time $t$ taken by the particle to move from $x=0$ to $x=5\text{cm}$ is given by
$5=5\sin \pi t$
or $1=\sin \pi t$
i.e $\pi t=\frac{\pi }{2}$ or $t=0.5\text{s}$