Question

A particle in S.H.M. is described by the displacement function $x\left(t\right)=acos(ax+\theta )$. 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

The correct option is B $\sqrt{2}cm$

$\begin{array}{c}Given,\\ Diplacementfunctionoftheparticleis,\\ x\left(t\right)=A\cos \left(\omega t+\varphi \right)\\ x\left(0\right)=1cmandv\left(0\right)=\pi \frac{cm}{\sec }\\ differentiatingw.r.tot,\\ v\left(t\right)=-A\left(\omega \right)\sin \left(\omega t+\phi \right)..........\left(1\right)\\ Att=0,\\ x\left(0\right)=A\omega \sin \left(\varphi \right)..........\left(2\right)\\ v\left(0\right)=-A\omega \sin \left(\varphi \right)............\left(3\right)\\ Dividing\left(3\right)by\left(2\right),\\ \frac{v\left(0\right)}{x\left(0\right)}=-\omega \tan \varphi \\ \Rightarrow \pi =-\omega \tan \varphi \\ \Rightarrow \tan \varphi =-1\\ i.e\varphi =\frac{3\pi }{4}or\frac{7\pi }{4}\\ \sin ce,att=0\\ x\left(0\right)=A\cos \left(\varphi \right)=1\\ \Rightarrow \cos \varphi ispositive.\\ Therefore,\varphi =\frac{7\pi }{4}\\ \therefore A\cos \left(\frac{7\pi }{4}\right)=1\\ \Rightarrow A=\sqrt{2}cm\end{array}$