A) \[\pi \]
B) \[2\pi \]
C) \[4\pi \]
D) \[6\pi \]
Correct Answer: C
Solution :
Given, \[4{{v}^{2}}=25-{{x}^{2}}\] Differentiating with respect to time t \[8v\frac{dv}{dt}=-2x\frac{dx}{dt}\] But \[\frac{dv}{dt}=a\](acceleration) and \[\frac{dx}{dt}=v\] \[\therefore \]\[8va=-2xv\]or \[a=-\frac{1}{4}x\] For SHM \[a=-{{\omega }^{2}}x\] \[\therefore \]\[{{\omega }^{2}}=\frac{1}{4}\]or \[\omega =\frac{1}{2}\] \[\therefore \]Period of Oscillation \[T=\frac{2\pi }{\omega }=4\pi \] Hence, the correction option is [c].You need to login to perform this action.
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