A)
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Correct Answer: D
Solution :
In simple harmonic motion, the force acting on the particle (restoring force) is given by \[\operatorname{F} = - kx\] (where k is positive constant) Now, \[F=-\frac{dU}{dx}\] \[\therefore \,\,\,\,-kx=-\frac{dU}{dx}\,\,\,\,\,or\,\,\,dU=kdx\] \[\therefore \,\,\,\,\,\,U(x)=\int_{0}^{x}{kdx}=\frac{1}{2}k{{x}^{2}}+C\] (C is constant of integration) In SHM the potential energy of the oscillator is zero at mean position, i.e., \[\operatorname{U}\left( 0 \right) =0\], Hence, \[\operatorname{c}= 0\] \[\therefore \,\,\,\,\,\,\,U\text{(}x)=\frac{1}{2}k{{x}^{2}}.\]which is equation of upward opening parabola.You need to login to perform this action.
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