Q. A particle in SHM is described by the displacement function $x(t) = A\ cos(\omega t + \phi$), $\omega = 2\pi/T.$ 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) $\sqrt{2}$ cm
D) 1 cm
Explanation
(c) $x(t) = A\cos{(\omega t + \phi)}$ …(i)
$\therefore\ \ \ 1 = A\cos{(\pi \times 0 + \phi)} = A\cos\phi$ …(ii)
Velocity$= \frac{d\lbrack x(t)\rbrack}{dt} = - A\omega\sin{(\omega t + \phi)}$
$\pi = - A \times \pi\sin{(0 + \phi)} = - \pi A\sin\phi - 1 = A\sin\phi$ …(iii)
Squaring and adding Eqs. (ii) and (iii), we have
$1 + 1 = A^{2}\left( \cos^{2}\phi + \sin^{2}\phi \right) = A^{2}$ or $A = \sqrt{2}$ cm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.