Question

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

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

Answer 1

Answer: (C)

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