The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = $$\dfrac{kx^2}{2}$$, 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 Figure. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Asked by Abhisek | 1 year ago |  73

##### Solution :-

Particle energy E = 1 J

K = 0.5 N m-1

K.E =$$\dfrac{1}{2}mv^2$$

Based on law of conservation of energy:

E = V + K

1 = $$\dfrac{1}{2}kx^2 + \dfrac{1}{2} mv^2$$

Velocity becomes zero when it turns back

1 = $$\dfrac{1}{2}kx^2$$

$$\dfrac{1}{2}\times 0.5x^2 = 1$$

X2 = 4

X = $$\pm2$$

Thus, on reaching x = $$\pm 2 m$$, the particle turns back.

Answered by Pragya Singh | 1 year ago

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