Question

A particle executes simple harmonic motion along a straight line with time period 12 sec. In what time the particle will cover half of the maximum displacement from the mean position?

Answer 1

Let the equation of SHM be
$x=Asin\omega t$
At half maximum displacement, $x=\frac{A}{2}$
$\frac{A}{2}=Asin\omega t\Rightarrow sin\omega t=\frac{1}{2}\Rightarrow \omega t=\frac{\pi }{6}$
$\omega =\frac{2\pi }{T}=\frac{2\pi }{12}$
$\Rightarrow \frac{2\pi }{12}t=\frac{\pi }{6}$
$\therefore t=15$