The potential energy function for a particle executing linear simple harmonic motion is given by U (x) = 
kx 2 , where k is the force constant. For k = 0.5 N m –1 , the graph of U(x) versus x is shown in figure. Show that a particle of total energy 1 J moving under this potential ‘turns back’ when it reaches x = ± 2m.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
At x = 0, total energy is in form of K.E. since U = 0
and it turns back when its K.E. = 0
So, total energy is in form of P.E.
U = – 
K
kx 2 = 1
x 2 = 1 × 2 × 2
x = ± 2m
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