Question

In a simple harmonic oscillator, at the mean position
A. Kinetic energy is minimum, potential energy is maximum
B. Both kinetic and potential energies are maximum
C. Kinetic energy is maximum, potential energy is minimum
D. Both kinetic and potential energies are minimum

Answer 1

Hint: in order to solve the question, we will first take the kinetic energy and potential energy in the kinetic energy we will substitute the value of velocity of SHM after then we will substitute the value of displacement at the mean position after then we will deduce the kinetic energy and potential energy.

Formula used:
$v = \omega \sqrt {{A^2} - {x^2}} $
Where, $v$ is the speed, $A$ is the amplitude, $\omega $ is the angular velocity and $x$ is the displacement of a particle performing S.H.M.
kinetic energy in S.H.M
$K.E = \dfrac{1}{2}m{v^2}$
Where, $M$ is mass and $V$ is the speed.
Potential energy in S.H.M
$P.E = \dfrac{1}{2}k{x^2}$
Where, $K$ is constant and $x$ is the displacement.

Complete step by step answer:
In the question we are asked about the variation in kinetic energy and potential energy when a simple harmonic oscillator is at its mean position. As $x$ is the displacement of a particle performing S.H.M so at the mean position value of displacement $x$ is zero.
$x = 0\,m$
The kinetic energy in the simple harmonic motion is
$K.E = \dfrac{1}{2}m{v^2}$
Now we will substitute the value of v in the kinetic energy equation
$v = \omega \sqrt {{A^2} - {x^2}} $

Substituting the value
$K.E = \dfrac{1}{2}m{\left( {\omega \sqrt {{A^2} - {x^2}} } \right)^2}$
Opening the bracket and square
$K.E = \dfrac{1}{2}m{\omega ^2}({A^2} - {x^2})$
Substitute the value of x = 0
$K.E = \dfrac{1}{2}m{\omega ^2}{A^2}$
The potential energy in the simple harmonic motion is
$P.E = \dfrac{1}{2}k{x^2}$
Substitute the value of x = 0
$\therefore P.E = 0$
Kinetic energy is maximum and potential value is zero that is minimum.

Hence, the correct option is C.

Note: Many of the students will make the mistake by not taking potential energy of SHM as zero because in some cases there are initial potential energy which doesn’t make it minimum but, in this case, potential energy is directly proportional to displacement which make it minimum and it have no initial energy.