Question

A simple harmonic motion along the $x$-axis has the following properties : amplitude $=0.5m$, the time to go from one extreme position to other is $2s$ and $x=0.3m$ at $t=0.5s$. The general equation of the simple harmonic motion is

• A. $x=(0.5m)\sin \left[\frac{\pi t}{2}+8^{\circ }\right]$
• B. $x=(0.5m)\sin \left[\frac{\pi t}{2}-8^{\circ }\right]$
• C. $x=(0.5m)\cos \left[\frac{\pi t}{2}+8^{\circ }\right]$
• D. $x=(0.5m)\cos \left[\frac{\pi t}{2}-8^{\circ }\right]$

Answer 1

Answer: (B)

Let equation of simple harmonic motion is
$x=A\sin (\omega t+\delta )$
It is given, $A=0.5m$ and
$\omega =\frac{2\pi }{T}=\frac{2\pi }{4}{s}^{-1}=\frac{\pi }{2}{s}^{-1}$
At $t=0.5s,x=0.3m$
so $0.3=0.5\sin (\omega t+\delta )$
$\Rightarrow \sin \left(\frac{\pi }{2}\times \frac{1}{2}+\delta \right)=\frac{3}{5}$
$\Rightarrow \frac{\pi }{4}+\delta =37^{\circ }\Rightarrow \delta =37^{\circ }-45^{\circ }=-8^{\circ }$
So, equation of motion is
$x=(0.5m)\sin \left[\frac{\pi t}{2}-8^{\circ }\right]$