### Video Transcript

A graph of a compression wave shows snapshots of the wave function for times π‘ equals 0.000 seconds in gold and π‘ equals 0.005 seconds in fuchsia. What is the wavelength of the compression wave? What is the maximum displacement of the compression wave? What is the speed of the compression wave? What is the period of the compression wave?

As we answer these questions, letβs let the Greek letter π represent wavelength. Weβll represent the maximum displacement as capital π΄ for amplitude. The wave speed weβll call π£. And the period of the wave weβll call capital π.

Now letβs look into these questions each in turn. First we want to solve for π, the wavelength of this wave.

Recall that the two curves we see in the diagram are the same wave, just at two different instants in time. We can work at either one to find the wavelength.

Letβs choose the fuchsia one. To find the wavelength of this wave, letβs pick an easily recognizable point on the curve. The wave trough at a value of π₯ equals five meters is a good starting point.

One wavelength from this point along the fuchsia curve will be the next trough over, at π₯ equals eleven meters. so π, the wavelength of this wave, is a positive value that is equal to the distance between these two points on the curve. So if we subtract five meters from 11 meters, we get a result of six meters.

If we assume that the values on our π₯- and π¦-axes are exact, then the only limit on the precision of this answer is the input values of time, which are given to three significant figures.

Two three significant figures, the wavelength of this curve is 6.00 meters. That is the distance from one trough of the curve to an adjacent trough or one crest of the curve to an adjacent crest.

Now we move on to solve for π΄, the amplitude of this wave. As we saw for the maximum displacement of the wave, also called its amplitude, that value is equal to the positive distance from the vertical midpoint of the wave, marked in our diagram by the π₯-axis and either the crest or trough of the wave.

For example, if we draw a vertical line from the π₯-axis to one of the wave crests, that distance is equal to the wave amplitude or its maximum displacement from equilibrium. As we look at the π¦-value of this point, we see that itβs equal to two millimeters.

The π¦-value of this point in space is equal to the amplitude of our wave or its maximum displacement, and that value is equal to 2.00 millimeters. That is the maximum displacement the wave makes from equilibrium.

Now letβs move on to solving for the wave speed represented by π£. We recall that, in general, the speed of an object is equal to the distance that object travels divided by the time itβs taken to travel that distance.

As we look at our diagram, we see the same wave, but with its position shown at two different instances in time. So when we apply this relationship for speed to our scenario, then the time value in the denominator will be equal to the difference in time between the snapshots of our wave.

In other words, the speed of our wave is equal to the distance it travels divided by 0.005 seconds minus 0.000 seconds. As we look at π, the distance the wave travels in that time, that will be equal to the distance between the crest of the fuchsia wave and an adjacent crest of the gold wave, for example, the horizontal distance between the point at π₯ equals two meters and the point at π₯ equals five meters.

So π equals 5.00 meters minus 2.00 meters. The speed of the wave then is 3.00 meters divided by 0.005 seconds, which is equal to 600 meters per second. Thatβs the speed at which the wave moves left to right in our diagram.

In the last part of our problem, weβre gonna solve for π‘, the period of the wave. To do that, letβs recall that the period π‘ is defined as one over the frequency of the wave, where the wave frequency is represented by a lowercase π.

Also, as it relates to wave speed, wave speed π£ is equal to the distance the wave travels divided by the time taken. And itβs also equal to the frequency of the wave multiplied by its wavelength.

Since weβve solved for the wavelength of the wave and the wave speed, we can use these two known values to solve for π, the frequency. To isolate π in the wave speed equation, we can divide both sides by π.

This results in π canceling out on the right side of our equation. And weβre left with a result that reads wave frequency π is equal to wave speed π£ divided by wavelength π.

When we insert 600 meters per second for π£ and 6.00 meters for π, we find that the frequency π of our wave is equal to 100 hertz, which is equal to 100 cycles per second.

Therefore, the period of our wave, capital π, which weβve said is equal to one divided by the frequency, is equal to one divided by 100 cycles per second, which is equal to 0.0100 seconds. Weβve now solved for the wavelength, maximum displacement, speed, and period of the wave shown in this diagram.