Course

ConceptPotential Energy of a Spring

#### My Profile

My Profile [view full profile]

why create a profile on shaalaa.com?

1. Inform you about time table of exam.

2. Inform you about new question papers.

3. New video tutorials information.

1. Inform you about time table of exam.

2. Inform you about new question papers.

3. New video tutorials information.

#### 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)