#### Question

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 = *0.5 N m^{–1}, 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

clickto share

#### Solution

You need to to view the solution

Is there an error in this question or solution?

#### Reference Material

Solution for question: The Potential Energy Function for a Particle Executing Linear Simple Harmonic Motion is Given By V(X) =Kx2/2, 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 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 concept: Potential Energy of a Spring. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)