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F 1 F T = ( y x ) x 1 and F 2 F T = ( y x ) x 2 .

The net force is on the small mass element can be written as

F net = F 1 + F 2 = F T [ ( y x ) x 2 ( y x ) x 1 ] .

Using Newton’s second law, the net force is equal to the mass times the acceleration. The linear density of the string μ is the mass per length of the string, and the mass of the portion of the string is μ Δ x ,

F T [ ( y x ) x 2 ( y x ) x 1 ] = Δ m a , F T [ ( y x ) x 2 ( y x ) x 1 ] = μ Δ x 2 y t 2 .

Dividing by F T Δ x and taking the limit as Δ x approaches zero,

[ ( y x ) x 2 ( y x ) x 1 ] Δ x = μ F T 2 y t 2 lim Δ x 0 [ ( y x ) x 2 ( y x ) x 1 ] Δ x = μ F T 2 y t 2 2 y x 2 = μ F T 2 y t 2 .

Recall that the linear wave equation is

2 y ( x , t ) x 2 = 1 v 2 2 y ( x , t ) t 2 .

Therefore,

1 v 2 = μ F T .

Solving for v , we see that the speed of the wave on a string depends on the tension and the linear density.

Speed of a wave on a string under tension

The speed of a pulse or wave on a string under tension can be found with the equation

| v | = F T μ

where F T is the tension in the string and μ is the mass per length of the string.

The wave speed of a guitar spring

On a six-string guitar, the high E string has a linear density of μ High E = 3.09 × 10 −4 kg/m and the low E string has a linear density of μ Low E = 5.78 × 10 −3 kg/m . (a) If the high E string is plucked, producing a wave in the string, what is the speed of the wave if the tension of the string is 56.40 N? (b) The linear density of the low E string is approximately 20 times greater than that of the high E string. For waves to travel through the low E string at the same wave speed as the high E, would the tension need to be larger or smaller than the high E string? What would be the approximate tension? (c) Calculate the tension of the low E string needed for the same wave speed.

Strategy

  1. The speed of the wave can be found from the linear density and the tension v = F T μ .
  2. From the equation v = F T μ , if the linear density is increased by a factor of almost 20, the tension would need to be increased by a factor of 20.
  3. Knowing the velocity and the linear density, the velocity equation can be solved for the force of tension F T = μ v 2 .

Solution

  1. Use the velocity equation to find the speed:
    v = F T μ = 56.40 N 3.09 × 10 −4 kg/m = 427.23 m/s .
  2. The tension would need to be increased by a factor of approximately 20. The tension would be slightly less than 1128 N.
  3. Use the velocity equation to find the actual tension:
    F T = μ v 2 = 5.78 × 10 −3 kg / m ( 427.23 m/s ) 2 = 1055.00 N .

    This solution is within 7 % of the approximation.

Significance

The standard notes of the six string (high E, B, G, D, A, low E) are tuned to vibrate at the fundamental frequencies (329.63 Hz, 246.94Hz, 196.00Hz, 146.83Hz, 110.00Hz, and 82.41Hz) when plucked. The frequencies depend on the speed of the waves on the string and the wavelength of the waves. The six strings have different linear densities and are “tuned” by changing the tensions in the strings. We will see in Interference of Waves that the wavelength depends on the length of the strings and the boundary conditions. To play notes other than the fundamental notes, the lengths of the strings are changed by pressing down on the strings.

Check Your Understanding The wave speed of a wave on a string depends on the tension and the linear mass density. If the tension is doubled, what happens to the speed of the waves on the string?

Since the speed of a wave on a taunt string is proportional to the square root of the tension divided by the linear density, the wave speed would increase by 2 .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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