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# Solution - 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 - Potential Energy of a Spring

ConceptPotential Energy of a Spring

#### Question

The potential energy function for a particle executing linear simple harmonic motion is given by V(x=kx2/2, where is the force constant of the oscillator. For k = 0.5 N m–1, the graph of V(x) versus 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 = ± 2 m

#### Solution

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#### APPEARS IN

NCERT Physics Textbook for Class 11 Part 1
Chapter 6: Work, Energy and Power
Q: 4 | Page no. 135

#### 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)
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