Question

A particle executing SHM is described by the displacement function $x\left(t\right)=Acos(\omega t+\varphi ),$ 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)

$\sqrt{2}cm$

$x=Acos(\omega t+\varphi )$ where A is amplitude.
At t = $0,x=1$ cm
$\therefore 1=Acos\varphi$ ....(i)
Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}\left(Acost\right(\omega t+\varphi \left)\right)=-A\omega sin(\omega t+\varphi )$
At t = $0,v=\pi cm{s}^{-1}$
$\therefore \pi =-A\omega sin\varphi OR\frac{\pi }{\omega }=-Asin\varphi$
$\therefore \omega =\pi s^{-1}$
$\therefore 1=-Asin\varphi$ ...(ii)

Squaring and adding (i) and (ii), we get
$A^{2}co{s}^2\varphi +A^{2}si{n}^2\varphi =2$
$A^2=2$
$\therefore A=\sqrt{2}cm$