How to Find Wave Speed from String Harmonics

A microphone beside a plucked guitar string might record frequency peaks at 196 Hz, 392 Hz, and 588 Hz. These are frequencies of sound travelling through the air, but they can also tell you how fast a wave travels along the string.

The key is to combine the harmonic frequency with the length of the vibrating string. For a string fixed at both ends, the wave speed is:

\[ \boxed{v=\frac{2Lf_n}{n}} \]

where:

• \(v\) is the transverse wave speed along the string, in \(\text{m s}^{-1}\).
• \(L\) is the vibrating length of the string, in metres.
• \(f_n\) is the frequency of the \(n\)th harmonic, in hertz.
• \(n\) is the harmonic number (1, 2, 3, and so on).

You do not need to measure the wavelength directly if you know the string length, harmonic number, and frequency. The challenge is identifying the correct harmonic number and understanding why the string and the surrounding air have different wave speeds.

01First, distinguish the string wave from the sound wave

Before calculating anything, there is an important distinction to understand.

When you pluck a string, two different types of wave are involved.

1. The transverse wave on the string

The particles making up the string move perpendicular to its length.

For a horizontal string, individual points move up and down while the wave disturbance travels along the string.

2. The longitudinal sound wave in the air

The vibrating string disturbs nearby air particles, producing alternating regions of compression and rarefaction.

Air particles oscillate backwards and forwards parallel to the direction in which the sound travels. They do not travel all the way from the guitar to your ear. In the basic wave model, they oscillate around their equilibrium positions.

Think of a stadium wave: the pattern travels around the stadium, but individual spectators mainly move locally. The analogy helps explain how a disturbance travels without transporting the particles along with it. Air molecules, of course, interact through pressure forces rather than performing a coordinated stadium routine.

Here is the important connection: the frequency of a sound component produced by the string matches the frequency of the corresponding string oscillation.

A microphone can therefore measure the frequency of the vibrating string indirectly.

However, the wavelength and speed are different in the two media:

\[ v_{\text{string}}=f\lambda_{\text{string}} \]

\[ c_{\text{air}}=f\lambda_{\text{air}} \]

Here, \(c_{\text{air}}\) is the speed of sound in air.

The frequency can be the same, but the wavelengths are not. When calculating the speed of a wave on the string, you must use the string’s wavelength, not the wavelength of the sound travelling through the air.

02How the string’s boundary conditions determine its wavelength

Imagine a string stretched tightly between two fixed points, such as the nut and bridge of a guitar.

Both ends must remain stationary. These stationary points are called nodes.

Between neighbouring nodes, the string oscillates. The point of maximum oscillation is called an antinode.

This restriction determines which standing-wave patterns can form.

For the fundamental frequency, one half-wavelength fits along the string.

For the second harmonic, two half-wavelengths fit.

For the third harmonic, three half-wavelengths fit.

This gives the general relationship:

\[ L=n\frac{\lambda_n}{2} \]

Rearranging:

\[ \boxed{\lambda_n=\frac{2L}{n}} \]

This is the relationship that makes the wave-speed calculation possible.

The number of loops in the standing-wave pattern equals the harmonic number. Each loop represents half a wavelength, not a full wavelength.

The fundamental is the first harmonic, with \(n=1\).

Higher harmonics have shorter wavelengths because more half-wavelengths must fit into the same vibrating length.

For an ideal string under constant tension, the wave speed stays the same across all these harmonics. Increasing the frequency does not make the wave travel faster along the string.

03The method: calculate wave speed from a harmonic

Now combine the wavelength relationship with the standard wave equation:

\[ v=f\lambda \]

For the \(n\)th harmonic:

\[ \lambda_n=\frac{2L}{n} \]

Therefore:

\[ v=f_n\left(\frac{2L}{n}\right) \]

Giving:

\[ \boxed{v=\frac{2Lf_n}{n}} \]

This is the formula to use when an HSC Physics question gives you the vibrating length and the frequency of a particular harmonic.

A quick decision process

Always check three things before substituting:

• Length: Use the vibrating length between the fixed nodes, not necessarily the total physical length of the string.
• Frequency: Use the frequency of the harmonic specified in the question.
• Harmonic number: A third harmonic means \(n=3\), not \(n=2\) or \(n=4\).

Keep length in metres and frequency in hertz. Your final wave speed will be in metres per second.

In diagrams, upward transverse displacement can be taken as positive. The calculated wave speed is a positive magnitude, rather than a signed displacement or velocity.

04The important trap: the lowest recorded frequency might not be the fundamental

Suppose a microphone records three peaks:

• \(392\text{ Hz}\)
• \(588\text{ Hz}\)
• \(784\text{ Hz}\)

A tempting prediction is that \(392\text{ Hz}\) must be the fundamental because it is the lowest recorded frequency.

That conclusion is not necessarily correct.

Imagine that an independent resonance investigation establishes that the fundamental frequency is actually \(196\text{ Hz}\).

The recorded peaks then correspond to:

\[ 392=2(196) \]

\[ 588=3(196) \]

\[ 784=4(196) \]

They are the second, third, and fourth harmonics.

The fundamental is present as an allowed natural mode, even though the microphone has not recorded a clear peak at that frequency.

Why might the fundamental be missing?

The amplitude of a harmonic depends on how the string is excited and how effectively its vibration produces detectable sound. The microphone, instrument body, and surrounding environment can also affect which frequencies are prominent.

A weak peak does not prove that the corresponding natural mode does not exist.

For example, suppose the string has a vibrating length of \(0.650\text{ m}\), and the \(392\text{ Hz}\) peak is independently identified as the second harmonic.

The correct calculation is:

\[ \begin{aligned} v&=\frac{2Lf_2}{2}\\ &=\frac{2(0.650)(392)}{2}\\ &=254.8\text{ m s}^{-1} \end{aligned} \]

If you incorrectly treat \(392\text{ Hz}\) as the fundamental, you would calculate:

\[ v=2(0.650)(392) \]

\[ v=509.6\text{ m s}^{-1} \]

That answer is exactly twice the correct value.

The mistake is not in the algebra. It is in assigning the wrong harmonic number.

There is another subtlety: frequency peaks alone do not always establish the fundamental uniquely. A set of peaks at 392, 588, and 784 Hz also consists of the fourth, sixth, and eighth multiples of 98 Hz.

You need additional evidence, such as a standing-wave observation or an independent measurement of the string’s properties, to settle the mode assignment.

05Assumptions and limitations

The formula \(v=2Lf_n/n\) relies on an idealised string model.

The main assumptions are:

• Both ends of the vibrating section behave as fixed nodes.
• The string has approximately uniform tension.
• The string has uniform linear mass density.
• The string is sufficiently flexible that its harmonics follow the ideal integer relationship.
• The recorded frequency corresponds to a vibration mode of the string, rather than an unrelated room or instrument resonance.

Real strings can have some stiffness, and real supports may move slightly. These effects can cause measured frequencies to deviate from the ideal harmonic sequence.

If you know the string’s tension and linear mass density, you can independently check your result using:

\[ v=\sqrt{\frac{T}{\mu}} \]

Here:

• \(T\) is the string tension in newtons (N).
• \(\mu\) is the linear mass density in kilograms per metre (\(\text{kg m}^{-1}\)).
• \(v\) is the transverse wave speed in \(\text{m s}^{-1}\).

If the two methods give substantially different answers, check your harmonic assignment, vibrating length, and assumptions about the string before accepting either result.

For a broader explanation of how tension, linear mass density, and standing-wave modes are related, see Standing Waves on Strings: Harmonics and Wave Speed.

If you are measuring wavelength directly using the distance between nodes in a practical investigation, see Standing-String Practical: Measure Wavelength and Wave Speed.

06Questions and solutions

Question 1

A string is fixed at both ends and has a vibrating length of \(0.60\text{ m}\). It produces a fifth harmonic with a frequency of \(250\text{ Hz}\).

Calculate:

(a) The wavelength of the fifth harmonic.

(b) The wave speed along the string.

Solution 1

The wavelength is \(0.240\text{ m}\), and the wave speed is \(60.0\text{ m s}^{-1}\).

Part (a): Calculate the wavelength.

The fifth harmonic has five half-wavelengths along the string.

Therefore, \(n=5\).

Use:

\[ \lambda_n=\frac{2L}{n} \]

Substitute:

\[ \begin{aligned} \lambda_5&=\frac{2(0.60)}{5}\\ &=\frac{1.20}{5}\\ &=0.240\text{ m} \end{aligned} \]

Part (b): Calculate the wave speed.

Use:

\[ v=f\lambda \]

Substitute the frequency and wavelength:

\[ \begin{aligned} v&=250\times0.240\\ &=60.0\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{v=60.0\text{ m s}^{-1}} \]

The string’s fifth harmonic has a relatively short wavelength, but the wave speed remains determined by the physical properties of the string.

Question 2

A string has a vibrating length of \(0.750\text{ m}\).

A frequency investigation identifies the following resonances:

• Second harmonic: \(210.0\text{ Hz}\)
• Fifth harmonic: \(526.0\text{ Hz}\)

The frequencies have been rounded to the nearest hertz.

(a) Calculate the wave speed using the second harmonic.

(b) Calculate the wave speed using the fifth harmonic.

(c) Determine the average of the two estimates and explain why they are slightly different.

Solution 2

The two estimates are \(157.5\text{ m s}^{-1}\) and \(157.8\text{ m s}^{-1}\). Their average is approximately \(157.7\text{ m s}^{-1}\).

Part (a): Use the second harmonic.

The given values are:

\[ L=0.750\text{ m},\quad n=2 \]

\[ f_2=210.0\text{ Hz} \]

Use:

\[ v=\frac{2Lf_n}{n} \]

Substitute:

\[ \begin{aligned} v_2&=\frac{2(0.750)(210.0)}{2}\\ &=157.5\text{ m s}^{-1} \end{aligned} \]

Part (b): Use the fifth harmonic.

Now:

\[ n=5,\quad f_5=526.0\text{ Hz} \]

Substitute:

\[ \begin{aligned} v_5&=\frac{2(0.750)(526.0)}{5}\\ &=157.8\text{ m s}^{-1} \end{aligned} \]

Part (c): Calculate the average and explain the difference.

Calculate:

\[ \begin{aligned} v_{\text{average}} &=\frac{157.5+157.8}{2}\\ &=157.65\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{v_{\text{average}}\approx157.7\text{ m s}^{-1}} \]

The difference between the two estimates is only \(0.3\text{ m s}^{-1}\).

This is consistent with the frequencies being rounded to the nearest hertz. The two results also overlap within their respective frequency-rounding uncertainties.

The wave speed has not increased between the second and fifth harmonics. The small difference comes from measurement precision, not a change in the physical properties of the string.

Question 3

A string has a vibrating length of \(0.72\text{ m}\). Its fundamental frequency is \(140\text{ Hz}\).

A movable fret is placed further along the string, reducing the vibrating length to \(0.54\text{ m}\).

Assume the tension and linear mass density remain unchanged.

Calculate:

(a) The original wave speed.

(b) The new fundamental frequency.

(c) The new third harmonic frequency.

Explain why the wave speed does not increase when the vibrating length is shortened.

Solution 3

The wave speed remains \(201.6\text{ m s}^{-1}\). The new fundamental frequency is approximately \(186.7\text{ Hz}\), and the new third harmonic is \(560\text{ Hz}\).

Part (a): Calculate the original wave speed.

The original fundamental is the first harmonic, so \(n=1\).

Use:

\[ v=2Lf_1 \]

Substitute:

\[ \begin{aligned} v&=2(0.72)(140)\\ &=201.6\text{ m s}^{-1} \end{aligned} \]

Part (b): Calculate the new fundamental frequency.

The tension and linear mass density have not changed.

Therefore, the wave speed remains constant.

Use:

\[ f_1=\frac{v}{2L} \]

The new vibrating length is \(0.54\text{ m}\).

\[ \begin{aligned} f_1&=\frac{201.6}{2(0.54)}\\ &=\frac{201.6}{1.08}\\ &=186.7\text{ Hz} \end{aligned} \]

Part (c): Calculate the new third harmonic.

For an ideal string:

\[ f_n=nf_1 \]

Therefore:

\[ \begin{aligned} f_3&=3(186.7)\\ &\approx560\text{ Hz} \end{aligned} \]

The third harmonic is:

\[ \boxed{f_3=560\text{ Hz}} \]

The important distinction is that shortening the string changes its allowed wavelengths and frequencies.

It does not automatically change the speed of waves travelling along it. Because the tension and linear mass density are unchanged, the wave speed remains constant.

Question 4

A string has a vibrating length of \(0.60\text{ m}\) and is oscillating in its third harmonic at \(450\text{ Hz}\).

(a) Calculate the transverse wavelength and wave speed on the string.

(b) Describe the motion of the air particles when the string produces sound.

(c) Explain why the wavelength of the sound in air cannot be calculated using the vibrating length of the string.

Solution 4

The wavelength on the string is \(0.40\text{ m}\), and its transverse wave speed is \(180\text{ m s}^{-1}\). The air particles oscillate parallel to the direction of sound propagation.

Part (a): Calculate the wavelength and wave speed.

The third harmonic has:

\[ n=3 \]

Use:

\[ \lambda_3=\frac{2L}{3} \]

Substitute:

\[ \begin{aligned} \lambda_3&=\frac{2(0.60)}{3}\\ &=0.40\text{ m} \end{aligned} \]

Now calculate the wave speed:

\[ v=f\lambda \]

\[ \begin{aligned} v&=450\times0.40\\ &=180\text{ m s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{v=180\text{ m s}^{-1}} \]

Part (b): Describe the air-particle motion.

The air particles oscillate backwards and forwards parallel to the direction in which the sound wave travels.

Their motion produces alternating compressions and rarefactions in the air.

The sound wave therefore travels longitudinally, while the wave on the string is transverse.

Part (c): Explain why the air wavelength is different.

The vibrating length of \(0.60\text{ m}\) describes the string, not the air through which the sound travels.

For the string, the third-harmonic wavelength is determined by the fixed-end boundary conditions:

\[ \lambda_{\text{string}}=\frac{2L}{3} \]

The sound wave in air has a different wavelength, determined by the speed of sound in air and its frequency:

\[ \lambda_{\text{air}}=\frac{c_{\text{air}}}{f} \]

Although both waves have the same frequency component of \(450\text{ Hz}\), they travel through different media at different speeds.

Using the air wavelength to calculate the string’s wave speed would therefore give an incorrect result.

Question 5

A uniform string is fixed at both ends and has a vibrating length of \(0.80\text{ m}\).

Its tension is \(52.8\text{ N}\), and its linear mass density is:

\[ \mu=7.60\times10^{-4}\text{ kg m}^{-1} \]

A microphone records three prominent frequency peaks:

\[ 330\text{ Hz},\quad495\text{ Hz},\quad660\text{ Hz} \]

There is no clear peak at \(165\text{ Hz}\).

Two students propose different explanations.

• Student A claims that \(330\text{ Hz}\) must be the fundamental because it is the lowest recorded frequency.
• Student B suggests that \(330\text{ Hz}\), \(495\text{ Hz}\), and \(660\text{ Hz}\) are the second, third, and fourth harmonics, respectively.

Assume the string follows the ideal flexible-string model, and the tension and linear mass density measurements are reliable.

(a) Calculate the transverse wave speed using the tension and linear mass density.

(b) Determine the theoretical fundamental frequency.

(c) Identify the harmonic number of each recorded peak.

(d) Evaluate the two students’ explanations. Explain why the missing \(165\text{ Hz}\) peak does not necessarily mean that the fundamental mode does not exist.

Solution 5

Student B is correct. The recorded peaks correspond to the second, third, and fourth harmonics, and the string’s wave speed is approximately \(264\text{ m s}^{-1}\).

The crucial point is that the lowest frequency detected by the microphone is not necessarily the fundamental frequency. The independent tension and linear mass density measurements allow us to resolve this ambiguity.

Part (a): Calculate the wave speed from tension and linear mass density.

For an ideal stretched string:

\[ v=\sqrt{\frac{T}{\mu}} \]

Substitute:

\[ v=\sqrt{\frac{52.8}{7.60\times10^{-4}}} \]

Therefore:

\[ \begin{aligned} v&=\sqrt{69473.7}\\ &=263.6\text{ m s}^{-1} \end{aligned} \]

So:

\[ \boxed{v\approx264\text{ m s}^{-1}} \]

This is an independent estimate of the wave speed. It does not rely on assigning harmonic numbers to the microphone peaks.

Part (b): Determine the theoretical fundamental frequency.

For a string fixed at both ends:

\[ f_1=\frac{v}{2L} \]

Substitute:

\[ \begin{aligned} f_1&=\frac{263.6}{2(0.80)}\\ &=\frac{263.6}{1.60}\\ &=164.75\text{ Hz} \end{aligned} \]

Therefore:

\[ \boxed{f_1\approx165\text{ Hz}} \]

The string’s fundamental frequency is approximately \(165\text{ Hz}\), even though the microphone does not show a clear peak at this frequency.

Part (c): Identify the harmonic numbers.

For an ideal string, the harmonic frequencies are integer multiples of the fundamental:

\[ f_n=nf_1 \]

Using \(f_1\approx164.75\text{ Hz}\):

Second harmonic:

\[ \begin{aligned} f_2&=2(164.75)\\ &=329.5\text{ Hz} \end{aligned} \]

Third harmonic:

\[ \begin{aligned} f_3&=3(164.75)\\ &=494.25\text{ Hz} \end{aligned} \]

Fourth harmonic:

\[ \begin{aligned} f_4&=4(164.75)\\ &=659.0\text{ Hz} \end{aligned} \]

Compare these predictions with the measured frequencies:

The agreement is very close.

Therefore, the recorded peaks correspond to:

\[ \boxed{330\text{ Hz}:n=2} \]

\[ \boxed{495\text{ Hz}:n=3} \]

\[ \boxed{660\text{ Hz}:n=4} \]

Part (d): Evaluate the two explanations.

Student A’s reasoning is tempting because \(330\text{ Hz}\) is the lowest frequency actually recorded.

However, the lowest recorded frequency is not necessarily the fundamental. A microphone measures the sound components that are sufficiently prominent to detect, not a guaranteed list of every natural mode.

If Student A incorrectly treats \(330\text{ Hz}\) as the fundamental, the calculated wave speed would be:

\[ \begin{aligned} v&=2Lf_1\\ &=2(0.80)(330)\\ &=528\text{ m s}^{-1} \end{aligned} \]

This is twice the speed predicted by the independently measured tension and linear mass density.

Student B’s explanation is consistent with the physical properties of the string.

There is one final subtlety worth noticing.

The three recorded peaks could also be interpreted as the fourth, sixth, and eighth harmonics of an \(82.5\text{ Hz}\) fundamental:

\[ 330=4(82.5) \]

\[ 495=6(82.5) \]

\[ 660=8(82.5) \]

The frequency spectrum alone cannot completely exclude this alternative.

However, an \(82.5\text{ Hz}\) fundamental would imply:

\[ \begin{aligned} v&=2Lf_1\\ &=2(0.80)(82.5)\\ &=132\text{ m s}^{-1} \end{aligned} \]

That contradicts the independently calculated wave speed of approximately \(264\text{ m s}^{-1}\).

The tension and linear mass density therefore resolve the ambiguity and establish the correct harmonic assignments.

The missing fundamental may simply have a weak amplitude or be poorly represented in the microphone recording. A natural mode can exist without appearing as a prominent peak in the measured sound spectrum.

The key decision in these problems is always the same: establish which harmonic you are using before calculating its wavelength. Once you know \(n\), the vibrating length, and the frequency, the wave speed follows directly. If the result seems inconsistent, compare it with \(v=\sqrt{T/\mu}\) rather than assuming that a higher harmonic must travel faster.