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 $1 cm$ 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 $v=\frac{d x}{d t}=A \frac{d}{d t} \cos (\omega t-\theta)$
$v=-A \omega \sin (\omega t+\theta)$
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 \,c m / s, $
$x=1 \,c m, $
$\omega=\pi s^{-1}$
$\therefore \pi=-\pi \sqrt{A^{2}-1}$
$\Rightarrow 1=A^{2}-1$
$\Rightarrow A=\sqrt{2}\, c m$