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Guide 13-1. Equations of Simple Harmonic Motion

Download this Excel file in order to experiment with changing the various parameters in order to see how that influences the graphs of position, velocity, and acceleration vs. time.

A. Phase relationships between position, velocity, and acceleration for an object in simple harmonic motion

See Section 13-2 of the text for more discussion of the equations.

For an object oscillating in SHM with angular frequency and released from rest at a position x = A, the position, velocity, and acceleration as a function of time are:

From these equations, one can see that the maximum values of the quantities are simply those for which or -1. They are:

Using the identity , the equation for velocity can be expressed as follows:

Using the identity , the equation for acceleration can be expressed as follows:

Now let's put the equations one below the other.

By expressing position, velocity, and acceleration as cosine functions, we see that velocity is shifted in phase by radians (90° or a quarter cycle) from position and acceleration is shifted in phase by by radians (180° or a half cycle) from position.

The graphs below show position, velocity, and acceleration all plotted on the same time scale. One can see that the velocity peaks a quarter of a cycle before the position and the acceleration peaks a quarter of a cycle before that (or a half cycle before the position.)  Let's write the equation for each graph. We first obtain the following information by reading the graphs.

T = 1.40 s               xmax = A = 1.60 m

Now we can calculate the following.

We can see that the latter two values agree with the corresponding graphs. We continue below the graphs and write the equations for x(t), v(t), and a(t).

Substituting values,

B. Equation of motion in the case of an equilibrium shift

In real-world situations, such as a mass oscillating vertically on a spring, the equilibrium position of the system may not correspond to a vertical position of 0 on a graph. See this animation as an example. Run the animation and note that the vertical path of the oscillation isn't centered on y = 0. Click on Show Graph to see the position vs. time graph. A screen capture of the graph is shown below.

The equation of the graph above is y = Acos(ωt) + yeq, where yeq is interpreted as a vertical shift due to the fact that the equilibrium position is not y = 0. The easiest way to determine the value of equilibrium shift from the graph is to average the maximum and minimum positions. For the graph above, this would be:

yeq = [0.14m + (-0.33m)]/2 = -0.095m

The amplitude of the motion is half the difference of the maximum and minimum positions:

A = [0.14m - (-0.33m)]/2 = 0.235m, or 0.24 to two decimal digits.

The period can be determined from the horizontal difference between 2 peaks. The angular frequency is then:

ω = 2π/T = 2π/(1.77s - 0.89s) = 7.1 s-1

Putting it all together, the equation of the position vs. time graph is:

y = (0.24m)cos[(7.1 s-1)t] - 0.095m

Let's check the result. At t = 1.6s, y = 0, according to the graph. The equation gives:

(0.24m)cos[(7.1 s-1)(1.6s)] - 0.095m = -0.01m

The result is the same as the precision of the vertical scale and is therefore attributable to sighting error in reading the scale.

We'll now go on to see how to deal with phase shifts.

C. Determining phase shifts

In section A, the phase of the position was 0. However, we can express position more generally using a phase shift. If we call the phase shift, then we write position as:  , where we're assuming an equilibrium shift of 0.  You may find it easier to interpret the phase shift as a time shift. The following sequence of equations shows how this arises.

In the above, we substitute for .  In like manner, we substitute for the phase shift, where ts is to be interpreted as a time shift. If we can determine ts from a plot of position vs. time, then we can calculate the phase shift using . For an example of how this is done, open this animation. Read the description. Then run the animation through one cycle of the motion. Stop the animation and click Show graph. The display should be something like the following (without the annotations).

The time shift is the difference in time between marks a and b where the red and blue graphs cross x = 0 going upward. (We could also have selected the two earlier points where the red and blue graphs cross x = 0 going downward.) Note at the bottom that the time grid is 0.1 s. So we can read the time shift as 5.1 grid divisions or 0.51 s. Now the period is the time from 0 to c. This is not quite 2 grid divisions less than the scale maximum of 3.0 s. We'll take it to be 2.82 s. Now that we have the time shift and the period, the phase difference is:

Enter that value into the input box for the Phase of Blue. Then reset and play the animation. Click Show graph. You should find that the two graphs coincide with each other very nearly.

There's another way to determine the phase shift using the animation. Play the animation from t = 0 to the point where the blue ball reaches the initial position of the red ball and with a velocity in the same direction as the red ball was moving at that time. Note the time. This is the time shift. Next measure the period of the motion, and you have what you need to calculate the phase shift.