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Frequencies and Wavelengths of Sound

Speed of Sound in a Gas

Doppler Effect for Sound

Doppler Effect Equations for Sound

Traveling Faster than Sound

Making Sounds with Musical Instruments

Sound or Music from a String

Equation for Sound from a String

Thickness and Material Factors in Equation for Sound from a Wire

Detecting Sound Waves

How Obstacles Affect Sound Waves

Sound Echoes

Sound Amplified Over Water

Reproducing Sounds

Beat Frequencies in Sound

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Explanation of the Equation for Sound from a String by Ron Kurtus - Succeed in Understanding Physics. Also refer to physics, wire, vibration, tension, length, mass, linear density, School for Champions. Copyright © Restrictions

Equation for Sound from a String

by Ron Kurtus (revised 1 February 2010)

When a string or wire is stretched between two posts and is plucked, it will vibrate and create a sound or musical note. The vibration of the string will create a fundamental frequency, which has its nodes at the end points. Harmonics, with nodes in regular positions along the length of the string, are also possible.

There is a general equation or formula that calculates the fundamental frequency, according to the tension, length, and mass of the string. Changing the various parameters results in changing the frequency of the vibration and thus the sound. You can also rearrange the equation to solve for the parameters.

Questions you may have include:

This lesson will answer those questions. There is a mini-quiz near the end of the lesson.

Useful tools: Metric-English Conversion | Scientific Calculator.

String frequency equation

The equation for the fundamental frequency of an ideal taut string is:

f = (1/2L)√(T/μ)

where

Note: Typically, tension would be in newtons, length in meters and linear density in kg/m, but those units are inconvenient for calculations with strings. Thus, the smaller units are used.

Linear density

Linear density is the mass per unit length: μ = m/L, where m is the mass of the string or wire in gm.

The reason μ is used instead of m/L is because when you use the equation to determine the frequency for a string of a different length, you must also adjust the mass to correspond to the different length. The situation where you change the length but keep the mass constant is seldom used.

Ideal string

The equation is actually an approximation for an ideal one-dimension string. Factors such as elasticity, material characteristics and diameter of the string or wire are not taken into account.

(See Thickness and Material Factors in Equation for Sound from a Wire for more information on that subject.)

The equation holds well if the amplitude of the string is small. The equation falls apart for strings plucked too vigorously.

Music

If the parameters of the string or wire—the length, tension and mass—are at certain values, the sound made from plucking the string will be a musical note that is pleasing to the ear. But if they are slightly different, the sound may not be musical and just be a sound.

Note that what is pleasing in one culture or nationality may not be considered musical in another culture.

Examples of changing the parameters

If the frequency for a given string—and the resulting sound—is a specific value, and you change one parameter of the string but keep everything else the same, the frequency will change accordingly.

Doubling the tension

If you double the tension, the frequency will go up 1.414 times the original frequency, provided all other parameters remain the same. Consider the frequency for a given configuration:

f1= (1/2L)√(T/μ)

If T is doubled, then the new frequency f2 is:

f2= (1/2L)√(2T/μ)

Your can take the square root of 2 to get:

f2= 1.414(1/2L)√(T/μ) = 1.414f1

Thus, if the frequency of the string is 500 Hz for a given configuration, and you double the tension of the string, the frequency goes up to 707 Hz.

f2 = 1.414f1= 1.414*500 Hz = 707 Hz

Shortening the length

If you shorten the length of the string by 1/2, while keeping all the other parameters constant, the frequency also goes up 2 times the original frequency.

f1 = (1/2L)√(T/μ)

Replace L with L/2:

f2= (2/2L])√(T/μ)

f2= 2f1

Again, if f1 = 400 Hz, then f2= 800 Hz.

Note: Since reducing the length by 1/2 also reduces the mass by 1/2, μ remains constant.

Reduce the mass

If you change the material of the string, reducing its mass by 1/2 but keeping the same length, the new frequency is 1.414 times the original frequency.

Substitute μ = m/L in f1 = (1/2L)√(T/μ):

f1 = (1/2L) √(TL/m)

Keeping other parameters constant, replace m with ½m:

f2= (1/2L)√(TL/½m)

f2 = (1/2L)√(2TL/m)

f2 = 1.414f1

Solving for other parameters

You can solve for the other parameters by squaring each side of the equation—or multiplying each item by itself—and rearranging them. Squaring a square root, gets rid of the square root sign and just leaves the number or variable.

Squaring both sides of the equation f = (1/2L)√(T/μ), results in:

f² = (T/μ)/4L² or

f² = T/4μL²

Find tension

Thus, if you know the frequency of the sound and the mass and length of the string, you can find the tension of the string:

T = 4μL²f²

Find length

If you know the tension and mass of the string and the frequency of the sound, you can find the length of the string:

L² = T/4μf²

L = (1/2f)√(T/μ)

Find mass

If you know the tension, the length of the string and the frequency of the sound, you can find the mass of the string:

m = T/4Lf²

Summary

A stretched string or wire will vibrate and create a sound or musical note when plucked. The vibration will be a fundamental frequency, according to the tension, length, and mass of the string. There is a general equation that calculates the frequency. Changing the parameters results in changing the frequency of the vibration and thus the sound.

Answers to Readers' Questions

See the Side Menu for more topics on Sound Waves


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Resources and references

Author's Credentials

The following resources provide information on this subject:

Websites

Vibrating String - Hyperphysics

Vibration of Stretched Strings - TutorVista.com

Vibrating string - Wikipedia

Physics Resources

Books

Top-rated books on Physical Science


Mini-quiz to check your understanding

1. Why is the equation f = (1/2L)√(T/μ) an approximation?

It doesn't take into account effects from the type of material

The square root is never exact

It depends on the length

2. If you increase the tension on the string 4 times, what happens to the fundamental frequency?

It goes down by 1/2

It doubles

It becomes 707 Hz

3. If you know the tension, length and frequency of the string, how do you determine its mass?

Enter the values to solve the equation m = T/4Lf²

Enter the values to solve the equation μ = m/L

You need more information to determine the mass

If you got all three correct, you are on your way to becoming a Champion in Physics. If you had problems, you had better look over the material again.


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