A) 1cm
B) \[\sqrt{2}\]cm
C) 2cm
D) 2.5cm
Correct Answer: B
Solution :
Key Idea: Rate of change of displacement gives velocity. Given, \[x=A\cos \,(\omega \,t+\theta )\] Velocity \[v=\frac{dx}{dt}=A\frac{d}{dt}\cos (\omega \,t-\theta )\] \[v=-A\,\omega \sin \,(\omega \,t+\theta )\] Using \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], we have \[v=-A\,\omega \sqrt{1-{{\cos }^{2}}(\omega \,t+\theta )}\] \[\therefore \] From equation\[x=A\cos \,(\omega \,t+\theta )\], we have \[v=-A\,\omega \sqrt{1-{{x}^{2}}/{{A}^{2}}}\] \[\Rightarrow \] \[v=-A\,\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\] Given, \[v=\pi \,\,cm/s,\,\,x=1\,cm,\,\,\omega =\pi \,{{s}^{-1}}\] \[\therefore \] \[\pi =-\pi \sqrt{{{A}^{2}}-1}\] \[\Rightarrow \] \[1={{A}^{2}}-1\] \[\Rightarrow \] \[A=\sqrt{2}\,cm\]
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