Question

A particle in S.H.M. is described by the displacement function$x(t) = a\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is 1 cm and its initial velocity is $\pi cm ⥂ / ⥂ s$. The angular frequency of the particle is $\pi rad/s$, then it’s amplitude is

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

Answer 1

Answer: (B)

$\sqrt{2}cm$

Answer 2

$x = a\cos(\omega t + \theta)$ ….(i)

and $v = \frac{dx}{dt} = - a\omega\sin(\omega t + \theta)$ ….(ii)

Given at $t = 0$, $x = 1cm$ and $v = \pi$ and $\omega = \pi$

Putting these values in equation (i) and (ii) we will get $\sin\theta = \frac{- 1}{a}$ and $\cos\theta = \frac{1}{A}$

⇒ $\sin^{2}\theta + \cos^{2}\theta = \left( - \frac{1}{a} \right)^{2} + \left( \frac{1}{a} \right)^{2}$⇒ $a = \sqrt{2}cm$