A) zero
B) 0.5 cm
C) 1 cm
D) 1.5 cm
Correct Answer: C
Solution :
Velocity \[\operatorname{v}=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\] and \[\operatorname{acceleration}\,\,=\,\,{{\omega }^{2}}x\] Given, \[\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\,\,=\,\,{{\omega }^{2}}x\] or \[\sqrt{{{A}^{2}}-{{x}^{2}}}\,\,=\,\,\omega x\] ... (i) Given, \[T=\frac{2\pi }{\sqrt{3}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\omega =\frac{2\pi }{T}=\sqrt{3}\] Substituting value of co in Eq. (i) We have \[\sqrt{{{A}^{2}}-{{x}^{2}}}\,\,=\,\,\sqrt{3}\,x\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\operatorname{A}=2x\] \[\operatorname{as} amplitude=\,\,\frac{Path\,\,length}{2}\,\,\times \,2\,cm\] \[\Rightarrow \,\,\,\,\,\,\,\,\,x =1 cm\]
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