Question

The potential energy of a particle oscillating along x-axis is given as U=20+(x-2)^(2) Here, U is in joules and x in meters. Total mechanical energy of the particle is 36J. (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle.

Answer 1

1)the given motion of particle is not a shm

              2)the mean position of the particle is 4m

               3)the maximum kinetic energy of the particle is 16J

The potential energy of the particle is given by

                             U=20+(x-2)^2

       the equation is in the form of a+x^2

      for a motion to be a SHM it should obey F∝-KX²

       the equation does not obey shm

          the max P.E=36J

          20+(x-2)^2=36⇒x=8,- 4

          the extream position s are at x=8,x=- 4

        the mean position is the mean of 8,-4 =4

        TOTAL ENERGY=K.E max+P.E max

     K.Emax=TOTAL ENERGY- P.E max

                    =36-20

                     =16J

Answer 2

Explanation:

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