Question

A particle in SHM is described by the displacement equation $x(t)=A\cos (\omega t+\theta )$. If the initial $(t=0)$ position of the particle is $1cm$ and its initial velocity is $\pi cm/s$, what is its amplitude? The angular frequency of the particle is $\pi s^{-1}$ :

• A. 1 cm
• B. $\sqrt{2}$ cm
• C. 2 cm
• D. 2.5 cm

Answer 1

Answer: (B)

Rate of change of displacement gives velocity.
Given, $x=A\cos (\omega t+\theta )$
Velocity $v=\frac{dx}{dt}=A\frac{d}{dt}\cos (\omega t-\theta )$
$v=-A\omega \sin (\omega t+\theta )$
Using ${\sin }^2\theta +{\cos }^2\theta =1$
we have $v=-A\omega \sqrt{1-{\cos }^2(\omega t+\theta )}$
$\therefore$ From equation $x=A\cos (\omega t+\theta )$
we have $v=-A\omega \sqrt{1-x^2}/A^2$
$\Rightarrow v=-A\omega \sqrt{A^2-x^2}$
Given, $v=\pi cm/s,$
$x=1cm,$
$\omega =\pi s^{-1}$
$\therefore \pi =-\pi \sqrt{A^2-1}$
$\Rightarrow 1=A^2-1$
$\Rightarrow A=\sqrt{2}cm$