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.

A neutron travelling with a velocity v and energy E collides elastically  with a nucleus of mass number A. Find the energy retained by the neutron  after collision. From Work, Energy and

Asked by Abhisek | 1 year ago |  73

1 Answer

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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