A) \[2\pi \sqrt{\frac{m}{(Ap)}}\]
B) \[2\pi \sqrt{\frac{m}{(A{{p}^{2}})}}\]
C) \[2\pi \sqrt{\frac{m}{A}}\]
D) \[\frac{1}{2\pi }\sqrt{\frac{Ap}{m}}\]
Correct Answer: B
Solution :
We are given that a particle of mass \[m\]is located in a one dimensional potential field and the potential energy is given by \[V(x)=A(1-\cos px).\] So, we can find the force experienced by the particle as \[F=-\frac{dV}{dx}=-Ap\sin \,px\] For small oscillations, we have \[F=-A{{p}^{2}}x\] Hence, the acceleration would be given by \[a=\frac{F}{m}=-\frac{A{{p}^{2}}}{m}x\] Also we know that \[a=\frac{F}{m}=-{{\omega }^{2}}x\] So, \[\omega =\sqrt{\frac{A{{p}^{2}}}{m}}\] or \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A{{p}^{2}}}}\]You need to login to perform this action.
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