Question

A particle executes simple harmonic motion of amplitude A along the x-axis. At $t=0$, the position of the particle is $x=\frac{A}{2}$ and it moves along the positive x-direction. Find the phase constant $\delta$, if the equation is written as $x=A\sin (\omega t+\delta )$ :

Answer 1

At the t=0 , the amplitude is $A/2$

So its position from horizontal axis is

$sin\delta =\frac{A/2}{A}=1/2$

$\delta =\frac{\pi }{6}$

So $x=Asin(\omega t+\frac{\pi }{6})$