Kinetic Energy in Simple Harmonic Motion

In most cases, it's almost intuitive that kinetic energy and velocity are related. For example, a car comes at you at \(\;5\;\mathrm{mph}\) while another comes at \(\;100\;\mathrm{mph}\). It's obvious which of the cars carries more kinetic energy. In a less intuitive example, we consider molecules in the air moving and oscillating. When we feel hot air, it means these molecules are moving fast, so they have high kinetic energy. When air is cold, it means these molecules are not moving so quickly, so they have low kinetic energy. In simple harmonic motion systems, the stiffness of the object exerting the restoring force on the oscillating object affects the kinetic energy of the system. In this article, we will understand how kinetic energy behaves in a system moving 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,

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

The Equation for the Kinetic Energy of a Spring

In a spring-mass system, when the spring decompresses, the spring does work on the object in order to move it. This work is equal to the potential energy stored in the spring, and during this moment the potential energy is converted to kinetic energy, so the object can move.

They are always interchanging, such that the total energy of the system is always constant. We will see maximum kinetic energy when the potential energy is at a minimum and vice versa.

In the case of a spring-mass system, we know the expression for the angular frequency is given by

$$\omega=\sqrt{\frac km},$$

where \(k\) is the spring constant that measures the stiffness of the spring in newtons per meter \((\frac{\mathrm N}{\mathrm m})\).

Now we can express the equation for the kinetic energy of a spring. The stiffer the spring, the higher the kinetic energy of the system. We will notice this as we substitute the value of \(\omega^2\) for a spring into the expression for the kinetic energy,

$$\begin{array}{rcl}K&=&\frac12m\omega^2\left(A^2-x^2\right),\\K&=&\frac12\cancel m\left(\frac k{\cancel m}\right)\left(A^2-x^2\right),\\K&=&\frac12k\left(A^2-x^2\right).\end{array}$$

Minimum and Maximum Kinetic Energy for a Spring

There are three moments in an oscillation cycle where the kinetic energy will be at its minimum. This occurs when the object is at maximum displacement. At these moments, the object's velocity is zero, as it changes its direction of motion. Also, during these moments, the displacement equals the amplitude. These moments correspond to \(t=0,\frac{\;T}{2,\;T}\):

$$\begin{array}{rcl}v_\min&=&\sqrt{\frac km}\sqrt{A^2-A^2},\\v_\min&=&0,\\K_\min&=&\frac12m{(0)}^2,\\K_\min&=&0\end{array}$$

On the other hand, kinetic energy will be at its maximum during two moments in an oscillation cycle. These moments occur when the object passes through the equilibrium position. These moments correspond to \(t=\frac T4,\;\frac{3T}4\):

$$\begin{array}{rcl}v_\max&=&\sqrt{\frac km}\sqrt{A^2-{(0)}^2},\\v_\max&=&A\sqrt{\frac km},\\K_\max&=&\frac12k(A^2-{(0)}^2),\\K_\max&=&\frac12kA.\end{array}$$