Question

A particle of mass m oscillates with simple harmonic motion between points \[{X_1}\] and \[{X_2}\], the equilibrium position being O. Its potential energy is plotted. It will be as given below in the graph:
A.
B.
C.
D.

Answer 1

Hint: If \[{X_1}\] and \[{X_2}\] represent two extreme positions then the potential energy(U) will be maximum at these points. At the point O which represents the equilibrium point then the potential energy(U) will be zero. If we plot a graph, it must show maximum values at \[{X_1}\] and \[{X_2}\] and also zero at O position.

Formula used:
Potential Energy in simple harmonic motion (SHM) is given as:
\[U = \dfrac{1}{2}k{x^2}\]
Where k is the force constant
x is a displacement

Complete step by step solution:
As we know that the for a particle of mass ‘m’ oscillates with simple harmonic motion (SHM) having Potential Energy,
\[U = \dfrac{1}{2}k{x^2}\]
This is in the form of parabolic equation which is in the form of \[y = {x^2}\]
At mean position, x=0 then the potential energy, U=0

Hence option A is the correct answer.

Note: Simple harmonic motion (SHM) is a type of an oscillatory motion in which the restoring force(F) is directly proportional to the displacement (x) of the particle from its mean position. The total energy of any particle while performing simple harmonic motion (SHM) is energy in simple harmonic motion. For the oscillatory motion, when a particle is at mean position it means at rest and when that particle reaches its extreme position then it comes to rest again. Thus, for the total energy in simple harmonic motion the kinetic energy and potential energy both must be calculated.