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Thickness and Material Factors in Equation for Sound from a Wire
by Ron Kurtus (1 February 2010)
A wire that is stretched between two posts and is plucked will vibrate and create a sound or musical note. The vibration of the wire will create a fundamental frequency, which has its nodes at the end points. There is an equation or formula to find the frequency of the sound as a function of the wire tension, length, diameter and density of the material, based on the Equation for Sound from a 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:
- What is the wire frequency equation?
- How can you change the parameters?
- What are equations for solving for the various parameters?
This lesson will answer those questions. There is a mini-quiz near the end of the lesson.
Useful tools: Metric-English Conversion | Scientific Calculator.
Wire frequency equation
The equation for the fundamental frequency of a taut wire as a function of tension, length, diameter and density of the wire material is:
f = (1/Ld)√(T/πδ)
where
- f is the frequency in hertz (Hz) or cycles per second
- L is the length of the wire in centimeters (cm)
- d is the diameter of the wire in cm
- T is the tension on the wire in gm-cm/s²
- π is the Greek letter pi = 3.14
- δ is the density of the wire in gm/cm³ (Greek letter small delta)
- √(T/πδ) is the square root of T divided by πδ
Derivation
The way this equation was derived is by starting with the general string equation:
f = (1/2L)√(T/μ)
(See Equation for Sound from a String for more information on the subject.)
Since the linear desity μ = m/L, the equation can be rewritten as:
f = (1/2L)√(TL/m)
where m is the mass of the string in grams (gm).
The mass of the wire can be written in terms of its density and volume:
m = δV
where V is the volume of the wire in cm³.
The volume is the length times the area of the cross section of the wire:
V = Lπd²/4
where πd²/4 is the area of the cross section in cm².
Substituting δV for m in f = (1/2L)√(TL/m), you get:
f = (1/2L)√(TL/δV)
Substituting Lπd²/4 for V, you get:
f = (1/2L)√(4TL/δLπd²)
Simplify to get:
f = (1/Ld)√(T/πδ)
Approximation
This equation is an approximation for an ideal one-dimension string. Adding the factors of diameter and density can reduce the accuracy a slight amount, provided the diameter is very small compared to the length.
The equation typically only applies to wires, because the density of a string or non-metal material has too much variation. You can select a wire material—such as copper—and find its density, while you seldom can find the density of a string.
Examples of changing the parameters
If the frequency from a given wire configuration has a specific value, you change one parameter of the wire, keeping everything else the same, to change the frequency. This can be stated as a percentage or proportional change.
Double the diameter
Changing the diameter of the wire changes its fundamental frequency. This can be seen on a guitar or piano, where the thicker wires have a lower frequency.
Let f1 be the initial frequency of the wire:
f1 = (1/Ld)√(T/πδ)
Keeping everything else the same, double the diameter d to become 2d:
f2 = (1/L2d)√(T/πδ)
f2 = f1/2
Thus, if f1 = 800 Hz, doubling the diameter of the wire will result in f2 = 400 Hz.
Changing the material
If you change the material of the wire, keeping everything else the same, you can compare the effect of the different densities. For example, the density of copper wire is 8.94 gm/cm³, while the density of steel wire is 7.8 gm/cm³.
Let fC be the frequency of the copper wire and fS be the frequency of the steel wire.
The frequency for a given copper wire is:
fC = (1/Ld)√(T/8.94π)
fC = 0.334(1/Ld)√(T/π)
Keeping everything else the same except the material, the frequency for a similar steel wire is:
fS = (1/Ld)√(T/7.8π)
fS = 0.358(1/Ld)√(T/π)
Comparing the two frequencies, you get:
fC = (0.334/0.358)fS
fC = 0.933fS
In other words, changing the wire from copper to the less-dense steel will result in a frequency that is 0.933 of the original. Thus, if the frequency of the copper wire was 800 Hz, the frequency of the steel wire will be 746 Hz.
Solving for other parameters
You can solve for the other parameters by squaring each side of the equation and rearranging them.
f = (1/Ld)√(T/πδ)
f² = (1/L²d²)(T/πδ)
Rearrange the variables:
f² = (T/πL²d²δ)
Solve for tension
You can rearrange the equation to solve for the tension:
T = πf²L²d²δ
Thus, if you know the other factors, you can find the tension on the wire.
Solve for length
You can rearrange the equation and take the square root of each side of the equation to solve for the length:
L² = T/f²d²πδ
L = √(T/f²d²πδ)
Simplify by taking the square root of 1/f²d²:
L = (1/fd )√(T/πδ)
Thus, if you know the other factors, you can find the length of the wire.
Solve for diameter
You can rearrange the equation and take the square root of each side of the equation to solve for the diameter of the wire:
d² = T/f²L²πδ
d = √(T/f²L²πδ)
d =(1/fL) √(T/πδ)
Thus, if you know the other factors, you can find the diameter of the wire.
Solve for density
You can rearrange the equation to solve for the density of the wire:
δ = T/f²L²d²π
Thus, if you know the other factors, you can find the density of the wire and thus its material.
Summary
A stretched wire that is between two posts and is plucked will vibrate at a fundamental frequency. The equation
or formula for the frequency of the sound as a function of the wire tension, length, diameter and density of the material
is
f = (1/Ld)√(T/πδ).
Changing the various parameters results in changing the frequency of the vibration. You can also rearrange the equation to solve for the various parameters.
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Resources
The following resources provide information on this subject:
Websites
Vibrating String - Hyperphysics
Vibration of Stretched Strings - TutorVista.com
Vibrating string - Wikipedia
Books
Top-rated
books on Physical Science
Mini-quiz to check your understanding
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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Thickness and Material Factors in Equation for Sound from a Wire