Computer Science, asked by ssandipgupta024, 3 months ago

6.4
The potential energy function for a
particle executing linear simple
harmonic motion is given by Vid =
kox /2, where k is the force constant
of the oscillator. For k = 0.5 Nm.
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.​

Answers

Answered by nirmalsunitha34
1

Answer:

The potential energy function for a particle executing linear simple harmonic motion is given by V(x)=kx

2

/2, where k is the force constant of the oscillator.

For k=0.5Nm

−1

, the graph of V(x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must turn back when it reaches x=±2m.

Explanation:

I Think this will help you

Answered by misrabarnali594
0

Answer:

hope it will be helpful to you

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