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  • The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 N m-1, the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of to

Q4  The potential energy function for a  particle executing linear simple harmonic motion is given by V ( x ) = k x^2 /2, where k is the force constant of the oscillator. For k = 0.5 Nm^{-1} the graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of  total energy 1 J moving under this potential must ‘turn back’ when it
reaches x = \pm 2

 

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The total energy of the particle is given by :

                                           E\ =\ K.E\ +\ P.E

or                                               =\ \frac{1}{2}mv^2\ +\ \frac{1}{2} kx^2

At the extreme position, the velocity of the object is zero thus its kinetic energy at that point is zero.

                                         E\ =\ \frac{1}{2} kx^2

or                                      1\ =\ \frac{1}{2} (0.5)x^2

or                                      x^2\ =\ 4

or                                      x\ =\ \pm \ 2

Hence the extreme position are \pm \ 2\ m. 

Posted by

Devendra Khairwa

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