Question

Two simple harmonic motions are denoted as $S_1$ and $S_2$ as below :
$S_1$: $x=6{\sin }^2\pi t$
$S_2$: $x(x-2)=\cos 3\pi t$
where $x$ is displacement (In cm) and $t$ is time (In second)

Answer 1

For$S_1:$
$x=6{\sin }^2\pi t$
$=3(1-\cos 2\pi t)$
$\Rightarrow (x-3)=-3\cos 2\pi t$
$\Rightarrow period=\frac{2\pi }{2\pi }=1second$
Amplitude$=3$
Centre of oscillation is at $x=3$
For$S_2:$
$x(x-2)=\cos 3\pi t$
$\Rightarrow (x-1{)}^2=1+\cos 3\pi t$
$=2co{s}^2\left(\frac{3\pi t}{2}\right)$
$\Rightarrow (x-1)=\sqrt{2}\cos \frac{3\pi t}{2}$
$\Rightarrow period=\frac{2\pi }{\frac{3\pi }{2}}=\frac{4}{3}s$
Amplitude$=\sqrt{2}$
Centre of oscillation is at $x=1$