Physics - SHM Question with Solution | TestHub

Given displacement equation is $Y=A\cos [\pi (t+\varphi )]$

At $t=0$, $Y=2$ cm

$A\cos (\pi \varphi )=2$

Velocity $v=\frac{dY}{dt}=-A\pi \sin [\pi (t+\varphi )]$

At $t=0$, $v=2\pi$ cm/s

$-A\pi \sin (\pi \varphi )=2\pi$

$-A\sin (\pi \varphi )=2$

Squaring and adding the two equations:

$A^2{\cos }^2(\pi \varphi )+A^2{\sin }^2(\pi \varphi )=2^2+(-2{)}^2$

$A^2=4+4=8$

$A=2\sqrt{2}$ cm

Answer: $2\sqrt{2}$