Kinetic Energy in Simple Harmonic Motion

Kinetic Energy of an Oscillator

The kinetic energy of an oscillator is associated with the energy required for its motion. The unit for kinetic energy is the joule \((\mathrm J)\) or newton-meters \((\mathrm N\;\mathrm m)\). It is important to notice that kinetic energy is a scalar quantity, not a vectorial quantity, meaning that it has magnitude, but it does not depend on any given direction. Velocity is a vector, but the magnitude of the velocity is a scalar quantity. To find the expression for the kinetic energy of an oscillator, we first need to find the velocity of an oscillator. We know that the kinetic energy of a particle is related to its mass and the square of its velocity, and is given by

$$K=\frac12mv^2.$$

In a previous article, we derived the expression for the potential energy of an oscillator:

$$U=\frac12\omega^2mx^2,$$

where \(\omega\) is the object's angular frequency in radians per second \((\frac{\mathrm{rad}}{\mathrm s})\).

We also stated that energy is conserved in simple harmonic motion. This means that at any two moments in an oscillation cycle, the sum of the kinetic and potential energies must be equal:

$$\begin{array}{rcl}K_i+U_i&=&K_f+U_f,\\\frac12mv_i^2+\frac12\omega^2mx_i^2&=&\frac12mv_f^2+\frac12\omega^2mx_f^2.\end{array}$$

Initially, we are at maximum displacement so, \(v_i=0,\;x_i=A,v_f=v,\;and\;x_f=x\). We substitute the values into the equation above and solve for the velocity:

$$\begin{array}{rcl}\frac12\omega^2mA^2&=&\frac12mv^2+\frac12\omega^2mx^2,\\v^2&=&\omega^2(A^2-x^2),\\v&=&\omega\sqrt{A^2-x^2}.\end{array}$$

Now that we know the expression for the velocity of the object undergoing simple harmonic motion, we can determine the equation for the kinetic energy of simple harmonic oscillators:

$$\begin{array}{rcl}K&=&\frac12mv^2,\\K&=&\frac12m\omega^2(A^2-x^2).\end{array}$$

As we can see in the above equation, there are many parameters in a system undertaking simple harmonic motion that can affect the kinetic energy. The kinetic energy is related to the mass of the oscillating object, its angular frequency, amplitude, and its position from the equilibrium point at any moment in time. The easiest way to prove this experimentally is by setting up a mass-spring system.

Another way to express the kinetic energy of an oscillator is using the definition of the position of an object in a simple harmonic motion system,

$$x=A\cos\left(\omega t+\phi\right).$$

We substitute the above equation in our expression for the kinetic energy,

$$\begin{array}{rcl}K&=&\frac12m\omega^2\left(A^2-A^2\cos^2\left(\omega t+\phi\right)\right),\\K&=&\frac12m\omega^2A^2\left(1-\cos^2\left(\omega t+\phi\right)\right).\end{array}$$

Where we use the trigonometric identity \(\cos^2\left(\theta\right)+\sin^2\left(\theta\right)=1\). So, we now have an expression for the kinetic energy which consists of a squared sine function. Notice that the squaring of the sine function means that the kinetic energy will always take positive values as we expect, even when the sine function itself is negative in value. The kinetic energy of a system undergoing simple harmonic motion is given by,

$$K=\frac12m\omega^2A^2\sin^2\left(\omega t+\phi\right).$$