Traveling waves

### Why is wavelength important?

When you make pulses in a medium, every pulse moves away from the source at the same constant speed. If you make pulses at regular time intervals, the result is a “train” of evenly spaced pulses. In technical terms, a periodic vibration creates a wave. The distance between one pulse and the next is the wavelength.

Wavelength explains a lot about how sound works- why tubas must be big, why low frequency sounds are hard to contain, why sonar can locate certain objects (and not others), how dead spots in auditoriums are formed, why instruments produce overtones- the list goes on. This section explains how wavelength is related to the frequency of the wave source and the properties of the medium that the wave travels in.

### Wavelength, speed and frequency

When a source with a certain frequency makes waves, each crest travels a certain distance before the next crest is formed. That distance is the wavelength. As a result, wavelength is controlled by just two factors: the frequency of the source and speed of waves.

If sound frequency is low, there’s a a long time between pulses and each pulse travels a long distance before the next one is made. The result is pulses that are spaced far apart- a long wavelength. Shorten the time delay between pulses, and the wavelength gets shorter. (Remember that all waves in the same medium travel at the same speed). The take-away point is that low frequency sources create long wavelength waves and high frequency sources create short wavelengths.

Wavelength is also affected by the medium, because waves travel at different speeds in different media. If the sound travels slow in a particular material, each crest travels only a short distance before the next crest is formed. If the same frequency source makes waves in a medium where sounds travel faster, each crest travels further (in the same amount of time), creating crests that are farther apart (a longer wavelength).

Animations are especially helpful for understanding the concepts in the last two paragraphs. Abbott’s Desmos simulation Pulse Train [1] shows two sources making waves in the same medium. Even though the sources have different frequencies, you can clearly see that both sets of waves travel at the same speed across the screen. You can also see that the high frequency source makes short wavelength waves.Mike Richardson’s youTube video [2] (4:49) uses the PhET simulation “Wave on a string” to show what happens when you change the frequency of a wave source on a string, If you want to try some of the things Richardson shows in the video, you can play with PhET simulation[3] yourself. You will need to check the box at the upper left of the simulation to get rid of the clamp and replace it with the open window- otherwise, you will be exploring standing waves rather than traveling waves.

Here’s the mathematical equation that relates wavelength to wave speed and source frequency:

[latex]\lambda = \dfrac{v }{f}[/latex]

The speed ([latex]v[/latex]) in the equation is the speed of wave in the medium. The frequency ([latex]f[/latex]) is the fundamental frequency of the source of the wave and [latex]\lambda[/latex] is the resulting wavelength.

### Where the equation comes from

The equation above comes directly applying the constant speed equation to a wave. Here’s the logic.Since all crests and troughs in a wave all travel at the same constant speed, crests and troughs (and all other points on a wave) travel according to the constant speed equation:

[latex]d = vt[/latex]

The next step is key: you have to recognize that, in the time it take the source to complete one full cycle, each ripple in a wave travels a distance of one wavelength. Since each crest (or trough) moves exactly one wavelength in the time it takes the source to complete one full cycle, you can replace [latex]d[/latex] with [latex]\lambda[/latex], provided you also replace [latex]t[/latex] with [latex]\tau[/latex]. The result is

[latex]\lambda= v\tau[/latex]

In words, wavelength equals the distance that a crest (or trough) travels in one period of the source.

Even though this equation is true and insightful, most people talk about sources in terms of frequency- not period. To express this equation in terms of frequency, replace [latex]\tau[/latex] with [latex]1/f[/latex]. When you do that, [latex]\lambda= v\tau[/latex] becomes

[latex]\lambda = \dfrac{v}{f}[/latex]

### Cause and effect

Many textbooks express the equation above as [latex]v=f\lambda[/latex]. I think that’s misleading.

Here’s why: the quantities [latex]\lambda[/latex], [latex]v[/latex] and [latex]f[/latex] are not on equal footing in the equation. Frequency and wave speed are causes. Wavelength is a response. Frequency is determined completely by the source- nothing else matters. Wave speed is completely determined by the properties of the medium- nothing else matters. Place a source that’s vibrating at a certain frequency in a particular medium, and the wave will automatically have a certain wavelength.

### Equation as a guide for thinking

If you know what you are doing, equations can be powerful guides for asking “what if” questions: What happens to wavelength if frequency increases? What happens to wavelength when a sound goes from water to air?

The key to “reading equations” is properly identifying which quantities change and which ones don’t. A quick glance at the equation [latex]v=f\lambda[/latex] might suggest that increasing the frequency of the source increases the wave speed. After all, using a bigger number for [latex]f[/latex] sound result in a larger number for [latex]v[/latex]. Yet, we know that conclusion is wrong- all sounds traveling in the same material travel at the same speed. What’s wrong with the math? The problem is we assumed [latex]\lambda[/latex] didn’t change. The reality is that [latex]v[/latex] cannot change (because the medium hasn’t changed), and [latex]\lambda[/latex] must change to compensate. If frequency increases, the wavelength must decrease in order for the wave speed to remain unchanged.

What happens to the wavelength of a sound that travels from water to air? The key to reading the equation for the answer is to identify which quantity is changing and which isn’t. As the sound moves from water to air, the speed decreases. (You can look this up; the key is to recognize that the speed of the wave must change, because the material in which the wave is traveling changes.) This suggests that frequency remains unchanged, and if you think microscopically it makes sense- particles vibrating on the far side of the border must match the vibrations of their neighbors on the near side of the boundary. Since [latex]f[/latex] remains constant and [latex]v[/latex] increases, the wavelength must increase to make the equation work.

### Stop to thinks

- Ripples moving on the surface of the water are produced by bobbing a long straight stick up and down on the surface of the water. The ripples can be changed by either 1) changing how often you bob the stick up and down or 2) changing how far up and down the stick goes each cycle (without changing how often you bob the stick up and down). Explain what you could do to the source to achieve each of the following goals:
- increase the amplitude of the waves
- shorten the wavelength of the waves
- increase the speed of the waves that are produced

- One sound in air has a wavelength of 10 cm. Another sound in air has a wavelength of 20 cm. Which sound has a higher frequency? Which sound travels faster?
- What happens to the speed of a wave as it travels from cold air to warm air? What happens to the frequency? What happens to the wavelength of the sound wave?

### Audible sounds in air

You can calculate the wavelengths of audible sound in air. Audible sounds in air have frequencies that range from roughly 20 Hz to 20 kHz. Not surprisingly, the wavelengths of audible sounds also vary widely. Assuming a speed of sound of 340 m/s,

For 20 Hz sound in air: [latex]\lambda = \dfrac{v}{f}= \dfrac{340 m/s}{20 Hz}=17 m[/latex]

For 20 kHz sound in air: [latex]\lambda = \dfrac{v}{f}= \dfrac{340 m/s}{20,000 Hz}=0.017 m=1.7 cm[/latex]

This calculation shows that wavelengths of sounds in air are distinctly human sized. The wavelengths range from roughly the diameter of a dime (for the highest frequencies) to roughly the length of a city bus (for the lowest frequencies). For comparison, the wavelengths of visible light are all far smaller than the thickness of a single human hair and have a very narrow range (from roughly 400 to 700 nm).

### Stop to think answers

- To increase amplitude, increase the distance that the stick travels each cycle (i.e. increase the amplitude of the source). To shorten the wavelength, increase how often you bob the stick (i.e. increase the frequency of the source). There is no way to change the speed of the ways by changing how you move the stick. The speed of the waves is determined by the properties of the water itself.
- The shorter wavelength sound has the higher frequency. Both sounds travel at the same speed.
- When the sound goes from cooler to warmer air, its speed increases (because sound travels faster in warmer air). The frequency doesn’t change (unless the source changes). Since speed increases and frequency is unchanged, the wavelength must increase. (Increasing the number for wave speed in the equation [latex]\lambda = v/f[/latex] without changing the number for frequency will lead to a bigger value for wavelength).

### Online resources

**Play with the PhET Sound simulation**(Java applet).[4] Adjust the frequency and watch how the wavelength changes. Click on the “Measure” tab at the top to add a ruler and stopwatch.

**Play with the PhET Waves on a String simulation** (HTML5).[5]. To make traveling waves, choose “No end” in the dialog box at the upper right.

### Image credit

Ripples on Loch Duich, in front of Eilean Donan castle. (by Guillaume Piolle; taken from Wikimedia Commons)[6]

- Abbott, D. (2018, July 27). Pulse Train. Retrieved from https://www.desmos.com/calculator/h7fy2cpthw ↵
- Richardson, M. (2013, February 26). Frequency versus wavelength. Retrieved from https://youtu.be/9UPnWfBYf28 ↵
- PhET Interactive Simulations. (2016). Wave on a string. Retrieved from https://phet.colorado.edu/en/simulation/wave-on-a-string ↵
- PhET (n.d.). Sound. Retrieved fromhttps://phet.colorado.edu/en/simulation/sound. ↵
- PhET (n.d.) Waves on a String. Retrieved from https://phet.colorado.edu/en/simulation/wave-on-a-string ↵
- Piolle, G. (2010, June 27). Ripples on Loch Duich, in front of Eilean Donan castle. Retrieved from https://commons.wikimedia.org/wiki/File:Eilean_Donan_castle_-_ripples.jpg ↵

## FAQs

### How do you find the frequency with wavelength and speed of sound? ›

If the wavelength and speed of a wave are known, these can be used to find the frequency of a wave using the equation **f = v λ** , where is the wavelength in meters and v is the speed of the wave in m/s. This also gives the frequency of the wave in Hertz.

**What do the frequency and wavelength of sound tell you? ›**

**The wavelength of a sound is the distance between adjacent identical parts of a wave**—for example, between adjacent compressions as illustrated in Figure 17.8. The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

**What is the relationship between speed frequency and wavelength? ›**

The relationship between the propagation speed, frequency, and wavelength is **v w = f λ** .

**How do you find the wavelength of the speed of sound? ›**

The relationship of the speed of sound vw, its frequency f, and its wavelength λ is given by **vw=fλ**, which is the same relationship given for all waves. In air, the speed of sound is related to air temperature T by vw=(331m/s)√T273K.