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 is tis motion is

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

Answer 1

Answer: (C)

$\sqrt{2}$ cm

Answer 2

$x = A\cos(\omega t) + \varphi$ where A is amplitude

At t = 0, x = 1 cm

$\therefore 1 = A\cos\varphi$

Velocity, $v = \frac{dx}{dt} = \frac{d}{dt}(A\cos(\omega t + \varphi)) = - A\omega\sin(\omega t + \varphi)$

At t = 0, $v = \pi cms^{- 1}$

$\therefore\pi = - A\omega\sin\varphi$ or $\frac{\pi}{\omega} = - A\sin\varphi$

$\because\omega = \pi s^{- 1}$

$\therefore 1 = - A\sin\varphi$

Squaring and adding (i) and (ii), we get

$A^{2}\cos^{2}\varphi + A^{2}\sin^{2}\varphi = 2$

$A^{2} = 2$ $(\because\sin^{2}\varphi + \cos^{2}\varphi = 1)$

$A = \sqrt{2}cm$