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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    A particle of mass m is located in a one dimensional potential field where potential energy is given by \[V(x)=A(1-\cos \,px),\]where A and p are constants. The period of small oscillations of the particle is

    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                            \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A{{p}^{2}}}}\]


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