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)

$\sqrt{2}$cm

Answer 2

Given, $v = \pi cm/sec,$ $x = 1cm$ and $\omega = \pi s^{- 1}$

using $v = \omega\sqrt{a^{2} - x^{2}}$⇒ $\pi = \pi\sqrt{a^{2} - 1}$

$\Rightarrow 1 = A^{2} - 1 \Rightarrow A = \sqrt{2}cm.$