Question

The potential energy function for a particle executing linear simple harmonic motion is given by V(x)=kx2/2, where k is force constant of the oscillator. For k=0.5Nm-1 the graph of V(x) versus ‘x’ is as shown in figure. Show that a particle of total energy 1J moving under this potential must ‘turn back’ when it reaches ±2m.

Answer 1

Total energy of the particle, E = 1 J
Force constant, k = 0.5 N m–1
Kinetic energy of the particle, K =1/2mv square(2)
According to the conservation law.
E = V + K.
1=1/2kx square(2)+1/2mv square(2)

At the moment of ‘turn back’, velocity (and hence K) becomes zero.
therefore 1=1/2kx(2)
1/2×0.5x(2) =1

x(2)=4
x=±2
Hence, the particle turns back when it reaches x =± 2 m

Answer 2

Answer:

Explanation:answer is +2 or - 2