Question

The total energy of the particle executing simple harmonic motion is:
A) $\alpha x$
B) $\alpha {x^2}$
C) \[{\text{Independent of}}\] $x$
D) $\alpha {x^{1/2}}$

Answer 1

Hint: In a simple harmonic oscillator, the energy is shared between the kinetic energy and potential energy. The total energy of the system is constant. In simple harmonic motion, there is a continuous interchange of kinetic energy and potential energy.

Formula Used:
We will be using the formulae of simple harmonic motion $x(t) = A\cos (\omega t + \phi )$, Potential energy $U = \dfrac{1}{2}k{x^2}$ , and Kinetic energy K.E$ = \dfrac{1}{2}m{v^{^2}}$.

Complete step by step answer:
To find the total energy of the particle executing simple harmonic motion.
We have to consider all forms of energy, to know the energy of a simple harmonic oscillator. In the simple harmonic motion of the mass and spring system, there are no dissipative forces. Hence the total energy of the system is equal to the sum of potential energy and kinetic energy. At the mean position of the simple harmonic motion, the total energy is kinetic energy and at the extreme position, the total energy in simple harmonic motion is purely potential energy and at other positions, the kinetic energy and potential energy are exchangeable.
For a particle in simple harmonic motion,
$x(t) = A\cos (\omega t + \phi )$
Therefore, the potential energy is given by,
$U = \dfrac{1}{2}k{x^2}$
Substitute $x$ value in potential energy, $U = \dfrac{1}{2}k{A^2}{\cos ^2}(\omega t + \phi )$
The kinetic energy is given by K.E $ = \dfrac{1}{2}m{v^{^2}}$
We know that, $v = \dfrac{{dx}}{{dt}}$
K.E= $\dfrac{1}{2}m{(\dfrac{{dx}}{{dt}})^2}$
$\dfrac{{dx}}{{dt}} = - A\omega \sin (\omega t + \phi )$
${(\dfrac{{dx}}{{dt}})^2} = {A^2}{\omega ^2}{\sin ^2}(\omega t + \phi )$
The kinetic energy becomes,
K.E$ = \dfrac{1}{2}m{A^2}{\omega ^2}{\sin ^2}(\omega t + \phi )$
We know that $\omega = \dfrac{{\sqrt k }}{{\sqrt m }}$
Where k is constant
   $K.E$ = $\dfrac{1}{2}k{A^2}{\sin ^2}(\omega t + \phi )$
The total energy is calculated by the sum of potential energy and kinetic energy
$E = U + K.E$
Using the trigonometric identities, $({\cos ^2}\phi + {\sin ^2}\phi = 1)$
$E = \dfrac{1}{2}k{A^2}{\cos ^2}(\omega t + \phi ) + \dfrac{1}{2}k{A^2}{\sin ^2}(\omega t + \phi )$
$E$$ = \dfrac{1}{2}k{A^{{2^{}}}}$
The total energy of a simple harmonic motion is equal to the square of the amplitude. The total energy of the system is constant

Therefore the total energy is independent of $x$; option (C).

Note: Every object possesses energy, it may be either moving or at rest. In the simple harmonic motion, the object moves to and fro along the same path. While traveling the same path, again and again, an object possesses energy in simple harmonic motion.