Lesson Explainer: Calculating Wave Motion Physics • 9th Grade

In this explainer, we will learn how to use the wave speed formula, 𝑠=𝑓𝜆, to calculate the movement of waves of different frequencies and wavelengths.

A wave is a type of disturbance that transfers energy from one point to another. Some waves need a medium to move through, like water waves or sound waves, which are vibrations of particles in the air. Other waves, such as light, can move through a vacuum.

One thing to remember when we talk about wave motion is that the motion of the wave refers to energy being transferred across some distance. Even when the wave is passing through a medium, the medium itself does not necessarily move very much. For example, individual water molecules move very little when a wave passes through them; they collide with other water molecules and pass the energy on but do not travel with the wave.

Two quantities that we use when discussing the motion of waves are the wavelength and amplitude. The wavelength of a wave is the distance needed for the wave to move through one complete cycle. The amplitude is the distance from the central or equilibrium point of a wave to the top of a crest or the bottom of a trough or the magnitude of its maximum displacement.

For a wave that is not changing, we can measure the amplitude from any peak or trough and measure the wavelength between any two successive points where the wave is in the same phase, as in the diagram below.

We can read the wavelength and amplitude of a wave from a displacement–distance graph.

The amplitude, which is equal to the magnitude of the maximum displacement, is 8 m.

For this wave, one complete cycle starting from a distance of 0 m consists of rising to its maximum displacement, falling through zero to its minimum, and then rising again back to zero. In the diagram below, the region highlighted in orange represents one cycle.

Note that it is not sufficient to simply return to the original displacement of 0 m. The wave must also be in the same phase as it was at the beginning of a cycle, that is, increasing its displacement.

The wave in the diagrams above takes 10 m to complete one cycle and so has a wavelength of 10 m. Note that this value is the same wherever we start in the wave’s cycle, provided we measure the distance taken to return to that same phase in the next cycle. We could have measured from the peak of the wave to the next peak, for example. When reading values from a graph, though, it is usually easier to choose a point where the wave crosses grid lines as the starting point.

So far, we have looked at waves on displacement–distance graphs. We can imagine these as being representations of a single instance of time, where we can see the wave’s phase changing with distance. Another way to represent waves is to consider the wave at a fixed point in space and measure its change in displacement over time. We can do this on a displacement–time graph.

On this graph, we can see that the wave takes a time of 1 s to complete one cycle. We say that this wave has a period of 1 s, where the period is defined as the amount of time taken for the wave to complete one cycle.

A more commonly used value is the frequency, which is defined as the number of cycles the wave completes in one second. If a wave has period 𝑝, then the frequency 𝑓=1𝑝. The unit of frequency is hertz, abbreviated Hz, where 1 Hz = 1 cycle/second. In the example above, 𝑝=1s, so we find the frequency from 𝑓=1𝑝=11=1.sHz

We could also read this directly from the graph, by noticing that the number of cycles completed in 1 s is 1, so the wave has a frequency of 1 Hz.

The following examples help us practice calculating the frequency of a wave.

Example 1: Understanding Wave Frequency

What is the frequency of the wave shown in the diagram?

Answer

The diagram is a displacement–time graph for a wave that starts with a displacement of 0 m at a time 0 s and oscillates between ±1.3m. From the diagram, we need to find the frequency of the wave.

Recall that the frequency of a wave is the number of cycles completed in 1 s. The diagram shows a time interval of 1 s, so we need to find the number of complete cycles shown on the diagram.

A complete cycle means that the wave must return to the same displacement and be in the same phase. Although it returns to a displacement of 0 m at around 0.25 s, this does not represent a complete cycle, because the displacement is decreasing here, whereas it was increasing at the start. The first cycle is completed at a time of 0.5 s. After a time of 1 s, the wave has completed two cycles. The frequency of the wave is therefore 2 Hz.

Example 2: Understanding Wave Frequency

What is the frequency of the wave shown in the diagram?

Answer

In this example, we have a displacement–time graph for a wave that takes 10 s to complete one cycle, and we need to find the frequency. Here, the period, 𝑝, of the wave is 10 s, so we can find the frequency, 𝑓, from 𝑓=1𝑝=110=0.1.sHz

So the frequency of the wave is 0.1 Hz.

We have discussed two properties of waves: the wavelength and the frequency. Another quantity we might find useful is the speed of the wave.

When we talk about the speed of a wave, we mean the speed at which a certain part of the wave travels, or propagates. Note that it is the energy, or the disturbance caused by the wave, that is traveling, not the material itself.

We can calculate the speed, 𝑠, of a wave from the frequency, 𝑓, and wavelength, 𝜆, as 𝑠=𝑓𝜆.

If we consider the units of 𝑓 and 𝜆 from the definitions of frequency and wavelength, we have [𝑠]=×.cyclestimedistancecycles

We have cycles on the numerator and denominator, so we can cancel those out and we have [𝑠]=,distancetime giving the usual unit for speed. If we have a wavelength measured in metres and a frequency in Hz (which is equivalent to 1s), we will therefore obtain a speed in a unit of metres per second (m/s).

To see this in practice, we will finish with some examples of using this equation.

Example 3: Calculating Wave Speed

A sound wave in a particular object has a frequency of 260 Hz and a wavelength of 2.5 m. At what speed does the sound wave propagate in that object, to the nearest metre per second?

Answer

In this example we are considering a sound wave. We are given the frequency, 𝑓=260Hz, and the wavelength, 2.5 m, and we need to calculate the speed of the wave.

Recall that wave speed, 𝑠, is related to frequency and wavelength by 𝑠=𝑓𝜆. Substituting the numbers in, we have 𝑠=𝑓𝜆=260×2.5=650/.Hzmms

So the sound wave is propagating at a speed of 650 m/s.

Example 4: Calculating Wave Frequency

A pier stretches from the coastline out into the sea a distance of 180 m. Waves on the sea move past the pier as they head toward the shore. The distance between the crests of the waves is 15 m, and the wave crests travel from the end of the pier to the shore in 24 seconds. What is the frequency of the waves?

Answer

Here, we are considering waves in water. We are told the distance between the crests of the waves is 15 m. This is the distance between successive points where the wave is in the same phase, so it is equal to the wavelength.

We are also told that the waves cover the distance from the end of the pier to shore, 180 m, in a time of 24 s. From this, we can calculate the waves’ speed as speeddistancetimemsms==18024=7.5/.

We now have the speed, 𝑠=7.5/ms, and the wavelength, 𝜆=15m, and we need to find the frequency.

Recall that speed, wavelength, and frequency are related by 𝑠=𝑓𝜆. We can find the frequency, 𝑓, by dividing both sides of this equation by 𝜆, which gives 𝑓=𝑠𝜆=7.5/15=0.5,msmHz and using the relation Hzs=1 to obtain the correct units for frequency.

The frequency of the wave is therefore 0.5 Hz.

Example 5: Understanding Wave Motion

The speed of the wave shown in the diagram is 460 m/s.

  1. What is the amplitude of the wave?
  2. What is the wavelength of the wave?
  3. What is the frequency of the wave?
  4. At what value of distance is the wave’s positive displacement equal to its amplitude?

Answer

Part 1

In this example, we are given a displacement–distance graph of a wave, and we are told that it has a speed of 460 m/s. The quantity we need to find is the amplitude.

Recall that the amplitude of a wave is the distance between its central or equilibrium position and its maximum displacement. In this example, the equilibrium displacement is 0 m, and the peak displacement is 3 m. The amplitude of the wave is therefore 3 m.

Part 2

Now, we are asked to find the wave’s wavelength. We can start from the first time it crosses the horizontal axis, at a distance of 0.5 m, at which point the displacement is decreasing. The next time the wave crosses the axis, at 2.5 m, the displacement is increasing, so the cycle is not complete until a distance of 4.5 m. The wavelength is therefore 4.50.5=4.0mmm.

Part 3

We now need to find the frequency of the wave. Since we have the wavelength, 𝜆=4m, and the speed, 𝑠=460/ms, we can find the frequency, 𝑓, from the relation 𝑠=𝑓𝜆. We first need to divide both sides of the equation by 𝜆: 𝑓=𝑠𝜆=460/4=115.msmHz

So the frequency of the wave is 115 Hz.

Part 4

Finally, we are asked to find the value of distance at which the wave’s positive displacement is equal to its amplitude. We found the amplitude above to be 3 m, so we need the value of distance at which the displacement is equal to 3 m. This occurs at a distance of 3.5 m.

Key Points

  • The amplitude of a wave is the magnitude of its maximum displacement from its equilibrium.
  • The wavelength is the distance needed for the wave to move through one complete cycle.
  • The period of a wave is the time taken to complete one cycle.
  • The frequency of a wave is the number of cycles completed in one second.
  • Frequency, 𝑓, is related to period, 𝑝, by 𝑓=1𝑝.
  • The unit of frequency is hertz, Hz, where 1 Hz = 1 cycle per second.
  • The speed, 𝑠, of a wave is related to frequency, 𝑓, and wavelength, 𝜆, as 𝑠=𝑓𝜆.

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