Question

A linear harmonic oscillator of force constant $2 \times {10^6}\,N{m^{ - 1}}$ and amplitude $0.01\,m$ has total mechanical energy of $160\,J$ . Which of the following statements are correct?
(A) The maximum PE of the particle is $100J$.
(B) The maximum KE of the particle is $100J$.
(C) The maximum PE of the particle is $160J$.
(D) The minimum PE of the particle is zero.

A. Only A is correct
B. Only B is correct
C. B and D are correct
D. B and C are correct

Answer 1

Hint: In order to solve this question we need to understand Simple harmonic motion. Simple harmonic motion is a motion in which applied force is directly proportional to displacement in the opposite direction. Also in Simple harmonic motion total mechanical energy is always conserved and constant, so at mean position or at equilibrium point potential energy is minimum and kinetic energy is maximum. At the extreme points kinetic energy is minimum and the potential energy is maximum.

Complete step by step answer:
We have given that, Force constant $k = 2 \times {10^6}\,N{m^{ - 1}}$.
Maximum amplitude is $A = 0.01\,m$.
Since we know at mean position Kinetic energy is maximum and it is given by,
$E = \dfrac{1}{2}k{A^2}$
Putting values we get,
$E = \dfrac{1}{2} \times (2 \times {10^6}) \times {({10^{ - 2}})^2}$
$\Rightarrow E = {10^{6 - 4}}J$
$\therefore E = 100\,J$
So maximum kinetic energy is $E = 100J$. Also we know at end points PE is maximum, so total mechanical energy is equivalent to potential energy.So Maximum PE of particle is $160\,J$.

So the correct option is D.

Note: It should be remembered that equilibrium point of simple harmonic motion is that point at which potential energy is minimum because exerted force is space derivative of potential so minimum potential energy corresponds minimum force and hence particle becomes stable at that point, so from energy conservation rule kinetic energy is maximum at equilibrium points.