Question

A particle in S.H.M. is described by the displacement function $x(t)=A\cos (\omega t+\varphi ),\omega =\frac{2\pi }{T}$ If the initial $(t=0)$ position of the particle is 1 cm, its initial velocity is n $cm{s}^{-1}$ and its angular frequency is $n{s}^{-1},$ then the amplitude of its motion is

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

Answer 1

Answer: (C)

: $x(t)=A\cos (\omega t+\varphi )$ where $\omega =\frac{2\pi }{T}$ At $t=0,x(t)=1$ $I=A\cos \varphi$ .... (i) velocity $v=\frac{dx}{dt}$ or $v=-A\sin (\omega t+\varphi )\times \omega$ or $v=-\omega A\sin (\omega t+\varphi )$ At $t=0,$ velocity $=\pi$ $\therefore$ $\pi =-\omega A\sin \varphi$ $\pi =-(\pi )A\sin \varphi$ or $A\sin \varphi =-1$ ....(ii) Square and add, $A^2{\cos }^2\varphi =A^2{\sin }^2\varphi ={(1)}^2+{(-1)}^2$ $A^2=2$ or $A=\sqrt{2}cm$