Question

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

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

Answer 1

Answer: (C)

$\sqrt2 \,cm$

Answer 2

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