Question

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

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

Answer 1

Answer: (B)

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