Simple harmonic motion

Simple harmonic motion is a unique form of oscillating motion where the restoring force is directly proportional to the displacement and acts in the opposite direction of that displacement.

Before we delve into a common application of simple harmonic motion, there are a few things to know:

A = amplitude, which is the distance of the object from equilibrium.

Equilibrium refers to the length of the spring when there are no forces acting on it.

x = the distance an object is from equilibrium.

k = spring constant

f = frequency, which is the number of cycles per second, which = 1/T.

T = period, which is the number of seconds per cycle, which = 1/f = 2pi√(m/k)

Simple harmonic motion can be represented through a graph with displacement on the y-axis and time on the x-axis.

Using this representation of simple harmonic motion, we can see that when the object is at its highest distance from equilibrium (height = +A or -A), the object has a velocity of zero, but a maximized acceleration (a = F/m = -kx/m, and since x is maximized at this moment, a is maximized).  Oppositely, when the objects distance from equilibrium is zero, then the velocity is maximized (the graph has the highest maximized slope) and the acceleration is minimized (x = 0, so a = 0).

Simple harmonic motion is most often assessed through springs.  Hooke’s Law is an equation that is essential to understanding how springs function: F = -kx.  The restoring force (F) is equal to the spring constant (k) multiplied by the displacement from equilibrium (x).  However, if the spring is aligned vertically and has a mass on it, then the equilibrium length of the spring is defined as the length of the spring with the mass on it.  An equation we can use to help us determine this length is mg = kx. This equation is derived from the free body diagram of a mass that is being held up by a spring:


Example Problem #1

If you have a 5 meter long spring with a spring constant of 50 Newton meters and attach it to the ceiling with a 3kg block hanging on it, what will the spring’s new length?

Answer: 5.588m

Explanation: Using the formula mg = kx, we can plug in values for m, g, and k: (3)(-9.8) = (50)(x).  Solving for x, we get that x = -.588.  Since we know the spring will get longer, we just need to know the magnitude of the displacement, which is .588m.  Adding this to the spring’s original length of 5m, we get 5.588 meters.

More Applications

Another application of simple harmonic motion is pendulums.  When working with pendulums, there are some important formulas to know:

T = 2pi√(l/g)

g = acceleration due to gravity (-9.8m/s^2)

l = length of string

The formula T = 2pi√(l/g) tells us that the time it takes for a pendulum to complete one full cycle (time it takes for it to reach the same spot twice) does NOT depend on how high you draw back the pendulum OR the mass of the object, but rather the length of the string and the acceleration due to gravity.

Example Problem #2:

If you have a pendulum on the moon (g = 1.625 m/s^2) and a string with a length of 1m, how long will it take for the pendulum to complete two cycles?

Answer: 9.86s

Explanation: For this problem, we are going to use the formula T = 2pi√(l/g).  Instead of using Earth’s acceleration due to gravity, we are going to have to use the moon’s (1.625 m/s^2).  Plugging in g = 1.625 m/s^2 and l = 1m, we get that T = 4.93s. Since the question asks for two cycles, however, we will multiply that value by two, getting 9.86s.