Question

A particle executing SHM is described by the displacement function $x(t)=\mathrm{Acos}(\omega t+$ $\varphi$ ), If the initial at $(t=0)$ position of the particle is $1cm$, its initial velocity is $\pi cm{s}^{-1}$ and its angular frequency is $\pi rad{s}^{-1}$, then the amplitude of its motion is:

• A. $\pi cm$
• B. $2cm$
• C. $\sqrt{2}cm$
• D. $1cm$

Answer 1

Answer: (C)

$x=A\cos (\omega t+\varphi )$ where $A$ is amplitude.
At $t=0,x=1cm$
$\therefore 1=A\cos \varphi$
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}(A\mathrm{cost}(\omega t+\varphi ))=-A\omega \sin (\omega t+\varphi )$
At $t=0,v=\pi cm{s}^{-1}$
$\therefore \pi =-A\omega \sin \varphi OR\frac{\pi }{\omega }=-A\sin \varphi$
$\therefore \omega =\pi s^{-1}$
$\therefore 1=-$ $Asin\varphi$
...(ii)
Squaring and adding (i) and (ii), we get
$A^2{\cos }^2\varphi +A^2{\sin }^2\varphi =2$
$A^2=2$
$\therefore A=\sqrt{2}cm$