Question

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

• A. $\pi\,cm$
• B. $2 \,cm$
• C. $ \sqrt2 \,cm$
• D. $1\,cm$

Answer 1

Answer: (C)

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