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 total energy 1 J moving under this potential must ‘turn back’ when it reaches x = � 2 m.


Given,

Total energy of the particle, E = 1 J


Force constant, k = 0.5 N m-1


The total energy is equal to the sum of kinetic energy and potential energy.


So, E = PE + K


E = kx2 + mv2


So, 1 = kx2 + mv2


When the velocity of the particle is zero i.e., at the turning point, the kinetic energy is zero.


1 = kx2


x2 = 2/k


x2 = 2/0.5


x2 = 4


x = �2


Hence, the particle turns back at x= �2.


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