You Are Given a Long Length of String and an Oscillator That Can Shake One End of the String

xvi Waves

16.iii Wave Speed on a Stretched Cord

Learning Objectives

By the end of this section, yous will exist able to:

Make up one's mind the factors that affect the speed of a wave on a string
Write a mathematical expression for the speed of a wave on a string and generalize these concepts for other media

The speed of a wave depends on the characteristics of the medium. For instance, in the case of a guitar, the strings vibrate to produce the sound. The speed of the waves on the strings, and the wavelength, determine the frequency of the sound produced. The strings on a guitar have different thickness merely may be made of similar textile. They have dissimilar linear densities, where the linear density is defined as the mass per length,

$\mu =\frac{\text{mass of string}}{\text{length of string}}=\frac{m}{l}.$

In this affiliate, we consider simply string with a constant linear density. If the linear density is constant, so the mass

$(\text{Δ}m)$

of a pocket-size length of string

$(\text{Δ}x)$

$\text{Δ}m=\mu \text{Δ}x.$

For instance, if the cord has a length of 2.00 thousand and a mass of 0.06 kg, then the linear density is

$\mu =\frac{0.06\,\text{kg}}{2.00\,\text{m}}=0.03\frac{\text{kg}}{\text{m}}.$

If a one.00-mm section is cut from the string, the mass of the one.00-mm length is

$\text{Δ}m=\mu \text{Δ}x=(0.03\frac{\text{kg}}{\text{m}})0.001\,\text{m}=3.00\,×\,{10}^{-5}\,\text{kg}.$

The guitar likewise has a method to change the tension of the strings. The tension of the strings is adjusted past turning spindles, called the tuning pegs, effectually which the strings are wrapped. For the guitar, the linear density of the cord and the tension in the string determine the speed of the waves in the cord and the frequency of the audio produced is proportional to the wave speed.

Moving ridge Speed on a String under Tension

To see how the speed of a wave on a string depends on the tension and the linear density, consider a pulse sent downwardly a taut cord ((Effigy)). When the taut string is at rest at the equilibrium position, the tension in the cord

${F}_{T}$

is constant. Consider a pocket-size chemical element of the cord with a mass equal to

$\text{Δ}m=\mu \text{Δ}x.$

The mass chemical element is at rest and in equilibrium and the strength of tension of either side of the mass element is equal and opposite.

Figure shows a section of a string with one portion highlighted. The length of the highlighted portion is labeled delta x. Two arrows from this portion point in opposite directions along the length of the string. These are labeled F subscript T. The highlighted portion is labeled delta m equal to mu delta x. — **Figure xvi.xiii** Mass element of a string kept taut with a tension
${F}_{T}$

. The mass element is in static equilibrium, and the force of tension acting on either side of the mass chemical element is equal in magnitude and reverse in direction.

If you pluck a string nether tension, a transverse wave moves in the positive x-management, as shown in (Figure). The mass element is modest just is enlarged in the figure to make it visible. The small mass element oscillates perpendicular to the wave motion every bit a result of the restoring forcefulness provided by the cord and does not move in the x-direction. The tension

${F}_{T}$

in the string, which acts in the positive and negative x-direction, is approximately constant and is independent of position and time.

Figure shows a pulse wave. Two arrows are shown along the upward slope of the wave, one pointing up and right, the other pointing down and left. These arrows, labeled F make angles theta 2 and theta 1 respectively with — **Figure xvi.xiv** A string under tension is plucked, causing a pulse to movement along the string in the positive ten-management.

Assume that the inclination of the displaced cord with respect to the horizontal centrality is small. The net forcefulness on the element of the string, interim parallel to the string, is the sum of the tension in the string and the restoring force. The x-components of the strength of tension cancel, then the net force is equal to the sum of the y-components of the force. The magnitude of the x-component of the strength is equal to the horizontal strength of tension of the string

${F}_{T}$

as shown in (Figure). To obtain the y-components of the force, note that

$\text{tan}\,{\theta }_{1}=\frac{\text{−}{F}_{1}}{{F}_{T}}$

and

$\text{tan}\,{\theta }_{2}=\frac{{F}_{2}}{{F}_{T}}.$

The

$\text{tan}\,\theta$

is equal to the gradient of a office at a point, which is equal to the partial derivative of y with respect to x at that point. Therefore,

$\frac{{F}_{1}}{{F}_{T}}$

is equal to the negative slope of the string at

${x}_{1}$

and

$\frac{{F}_{2}}{{F}_{T}}$

is equal to the gradient of the string at

${x}_{2}:$

$\frac{{F}_{1}}{{F}_{T}}=\text{−}{(\frac{\partial y}{\partial x})}_{{x}_{1}}\,\text{and}\,\frac{{F}_{2}}{{F}_{T}}=\text{−}{(\frac{\partial y}{\partial x})}_{{x}_{2}}.$

The net strength is on the small mass element tin exist written as

${F}_{\text{net}}={F}_{1}+{F}_{2}={F}_{T}[{(\frac{\partial y}{\partial x})}_{{x}_{2}}-{(\frac{\partial y}{\partial x})}_{{x}_{1}}].$

Using Newton's 2nd law, the net force is equal to the mass times the dispatch. The linear density of the string

$\mu$

is the mass per length of the cord, and the mass of the portion of the string is

$\mu \text{Δ}x$

$\begin{array}{cc} {F}_{T}[{(\frac{\partial y}{\partial x})}_{{x}_{2}}-{(\frac{\partial y}{\partial x})}_{{x}_{1}}]=\text{Δ}ma,\hfill \\ {F}_{T}[{(\frac{\partial y}{\partial x})}_{{x}_{2}}-{(\frac{\partial y}{\partial x})}_{{x}_{1}}]=\mu \text{Δ}x\frac{{\partial }^{2}y}{\partial {t}^{2}}.\hfill \end{array}$

Dividing by

${F}_{T}\text{Δ}x$

and taking the limit as

$\text{Δ}x$

approaches zero,

$\begin{array}{ccc}\hfill \frac{[{(\frac{\partial y}{\partial x})}_{{x}_{2}}-{(\frac{\partial y}{\partial x})}_{{x}_{1}}]}{\text{Δ}x}& =\hfill & \frac{\mu }{{F}_{T}}\,\frac{{\partial }^{2}y}{\partial {t}^{2}}\hfill \\ \hfill \underset{\text{Δ}x\to 0}{\text{lim}}\frac{[{(\frac{\partial y}{\partial x})}_{{x}_{2}}-{(\frac{\partial y}{\partial x})}_{{x}_{1}}]}{\text{Δ}x}& =\hfill & \frac{\mu }{{F}_{T}}\,\frac{{\partial }^{2}y}{\partial {t}^{2}}\hfill \\ \hfill \frac{{\partial }^{2}y}{\partial {x}^{2}}=\frac{\mu }{{F}_{T}}\,\frac{{\partial }^{2}y}{\partial {t}^{2}}.\end{array}$

Recall that the linear wave equation is

$\frac{{\partial }^{2}y(x,t)}{\partial {x}^{2}}=\frac{1}{{v}^{2}}\,\frac{{\partial }^{2}y(x,t)}{\partial {t}^{2}}.$

Therefore,

$\frac{1}{{v}^{2}}=\frac{\mu }{{F}_{T}}.$

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

Speed of a Wave on a String Nether Tension

The speed of a pulse or moving ridge on a cord under tension tin can be constitute with the equation

$|v|=\sqrt{\frac{{F}_{T}}{\mu }}$

where

${F}_{T}$

is the tension in the string and

$\mu$

is the mass per length of the string.

Instance

The Wave Speed of a Guitar Spring

On a six-cord guitar, the loftier Due east string has a linear density of

${\mu }_{\text{High E}}=3.09\,×\,{10}^{-4}\,\text{kg/m}$

and the low E string has a linear density of

${\mu }_{\text{Low E}}=5.78\,×\,{10}^{-3}\,\text{kg/m}.$

(a) If the high Eastward 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 East string is approximately 20 times greater than that of the high E string. For waves to travel through the low East string at the same moving ridge speed every bit the high Eastward, would the tension need to be larger or smaller than the high E string? What would exist the approximate tension? (c) Calculate the tension of the depression East string needed for the same wave speed.

Strategy

The speed of the moving ridge tin be institute from the linear density and the tension
$v=\sqrt{\frac{{F}_{T}}{\mu }}.$
From the equation
$v=\sqrt{\frac{{F}_{T}}{\mu }},$

if the linear density is increased by a factor of nigh 20, the tension would need to exist increased by a gene of 20.
Knowing the velocity and the linear density, the velocity equation can be solved for the forcefulness of tension
${F}_{T}=\mu {v}^{2}.$

Solution

Use the velocity equation to find the speed:
$v=\sqrt{\frac{{F}_{T}}{\mu }}=\sqrt{\frac{56.40\,\text{N}}{3.09\,×\,{10}^{-4}\,\text{kg/m}}}=427.23\,\text{m/s}.$
The tension would demand to be increased by a cistron of approximately 20. The tension would exist slightly less than 1128 N.

Use the velocity equation to find the bodily tension:

${F}_{T}=\mu {v}^{2}=5.78\,×\,{10}^{-3}\text{kg}\text{/}\text{m}{(427.23\,\text{m/s})}^{2}=1055.00\,\text{N}.$

This solution is within

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of the approximation.

Significance

The standard notes of the six string (high E, B, K, D, A, depression East) 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 cord and the wavelength of the waves. The half-dozen strings take 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 inverse by pressing downwards on the strings.

Check Your Understanding

The wave speed of a moving ridge 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 cord?

[reveal-answer q="fs-id1165038278014″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165038278014″]

Since the speed of a moving ridge on a taunt string is proportional to the foursquare root of the tension divided by the linear density, the wave speed would increment by

$\sqrt{2}.$

[/hidden-reply]

Speed of Pinch Waves in a Fluid

The speed of a moving ridge on a string depends on the square root of the tension divided past the mass per length, the linear density. In full general, the speed of a wave through a medium depends on the elastic belongings of the medium and the inertial property of the medium.

$|v|=\sqrt{\frac{\text{elastic property}}{\text{inertial property}}}$

The elastic property describes the tendency of the particles of the medium to render to their initial position when perturbed. The inertial property describes the trend of the particle to resist changes in velocity.

The speed of a longitudinal wave through a liquid or gas depends on the density of the fluid and the majority modulus of the fluid,

$v=\sqrt{\frac{Β}{\rho }}.$

Here the bulk modulus is defined as

$Β=-\frac{\vartriangle P}{\frac{{\vartriangle V}}{{V}_{0}}},$

where

$\text{Δ}P$

is the change in the pressure and the denominator is the ratio of the change in volume to the initial volume, and

$\rho \equiv \frac{m}{V}$

is the mass per unit volume. For instance, sound is a mechanical wave that travels through a fluid or a solid. The speed of sound in air with an atmospheric pressure of

$1.013\,×\,{10}^{5}\,\text{Pa}$

and a temperature of

$20\text{°}\text{C}$

${v}_{\text{s}}\approx 343.00\,\text{m/s}.$

Because the density depends on temperature, the speed of sound in air depends on the temperature of the air. This will be discussed in particular in Sound.

Summary

The speed of a wave on a string depends on the linear density of the string and the tension in the string. The linear density is mass per unit of measurement length of the cord.
In general, the speed of a wave depends on the square root of the ratio of the elastic property to the inertial property of the medium.
The speed of a wave through a fluid is equal to the square root of the ratio of the bulk modulus of the fluid to the density of the fluid.
The speed of sound through air at
$T=20\text{°}\text{C}$

is approximately

${v}_{\text{s}}=343.00\,\text{m/s}.$

Conceptual Questions

If the tension in a string were increased by a gene of four, by what gene would the moving ridge speed of a wave on the string increase?

[reveal-answer q="936911″]Show Solution[/reveal-answer]
[hidden-answer a="936911″]The wave speed is proportional to the square root of the tension, so the speed is doubled.[/subconscious-answer]

Does a sound wave motility faster in seawater or fresh water, if both the sea h2o and fresh water are at the same temperature and the audio wave moves near the surface?

$({\rho }_{\text{w}}\approx 1000\frac{\text{kg}}{{\text{m}}^{3}},{\rho }_{\text{s}}\approx 1030\frac{\text{kg}}{{\text{m}}^{3}},{B}_{\text{w}}=2.15\,×\,{10}^{9}\,\text{Pa},$

${B}_{\text{s}}=2.34\,×\,{10}^{9}\,\text{Pa})$

Guitars have strings of different linear mass density. If the everyman density string and the highest density cord are under the same tension, which cord would support waves with the higher wave speed?

[reveal-answer q="177839″]Show Solution[/reveal-reply]
[hidden-reply a="177839″]Since the speed of a wave on a string is inversely proportional to the square root of the linear mass density, the speed would exist college in the low linear mass density of the string.[/hidden-answer]

Shown below are three waves that were sent down a string at different times. The tension in the string remains constant. (a) Rank the waves from the smallest wavelength to the largest wavelength. (b) Rank the waves from the lowest frequency to the highest frequency.

Figure shows three waves labeled A, B and C on the same graph. All have their equilibrium positions on the x axis. Wave A has amplitude of 4 units. It has crests at x = 1.5 and x = 7.5. Wave B has amplitude of 3 units. It has a crest at x = 2 and a trough at x = 6. Wave C has amplitude of 2 units. It has crests at x = 1 and x = 5.

Electrical power lines continued by two utility poles are sometimes heard to hum when driven into oscillation past the wind. The speed of the waves on the power lines depend on the tension. What provides the tension in the power lines?

[reveal-answer q="960058″]Bear witness Solution[/reveal-answer]
[hidden-respond a="960058″]The tension in the wire is due to the weight of the electric ability cable.[/hidden-answer]

Ii strings, one with a low mass density and ane with a loftier linear density are spliced together. The college density cease is tied to a lab mail service and a student holds the free terminate of the low-mass density string. The educatee gives the string a flip and sends a pulse downward the strings. If the tension is the same in both strings, does the pulse travel at the same wave velocity in both strings? If non, where does information technology travel faster, in the low density string or the high density cord?

Problems

Transverse waves are sent along a 5.00-k-long cord with a speed of thirty.00 m/due south. The string is under a tension of x.00 Due north. What is the mass of the cord?

A copper wire has a density of

$\rho =8920\,{\text{kg/m}}^{3},$

a radius of 1.twenty mm, and a length 50. The wire is held under a tension of 10.00 N. Transverse waves are sent down the wire. (a) What is the linear mass density of the wire? (b) What is the speed of the waves through the wire?
[reveal-reply q="492763″]Bear witness Solution[/reveal-answer]
[subconscious-answer a="492763″]a.

$\mu =0.040\,\text{kg/m;}$

$v=15.75\,\text{m/s}$

[/hidden-answer]

A piano wire has a linear mass density of

$\mu =4.95\,×\,{10}^{-3}\,\text{kg/m}.$

Under what tension must the string be kept to produce waves with a wave speed of 500.00 m/s?

A string with a linear mass density of

$\mu =0.0060\,\text{kg/m}$

is tied to the ceiling. A 20-kg mass is tied to the free end of the cord. The string is plucked, sending a pulse downwards the cord. Estimate the speed of the pulse as it moves downwards the string.

[reveal-answer q="fs-id1165037050239″]Show Solution[/reveal-respond]

[hidden-answer a="fs-id1165037050239″]

$v=180\,\text{m/s}$

[/hidden-answer]

A cord has a linear mass density of

$\mu =0.0075\,\text{kg/m}$

and a length of iii meters. The cord is plucked and it takes 0.20 s for the pulse to achieve the end of the cord. What is the tension of the string?

A string is 3.00 g long with a mass of 5.00 g. The string is held taut with a tension of 500.00 Due north practical to the cord. A pulse is sent down the string. How long does it take the pulse to travel the three.00 g of the string?

[reveal-answer q="439036″]Bear witness Solution[/reveal-answer]
[hidden-answer a="439036″]

$v=547.723\,\text{m/s},\,\text{Δ}t=5.48\,\text{ms}$

[/hidden-respond]

Two strings are attached to poles, all the same the kickoff string is twice as long as the second. If both strings have the same tension and mu, what is the ratio of the speed of the pulse of the moving ridge from the first string to the second cord?

Ii strings are attached to poles, still the get-go string is twice the linear mass density mu of the 2d. If both strings have the same tension, what is the ratio of the speed of the pulse of the moving ridge from the outset string to the second cord?

[reveal-answer q="fs-id1165037203500″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1165037203500″]

0.707

[/hidden-answer]

Transverse waves travel through a string where the tension equals vii.00 N with a speed of twenty.00 m/south. What tension would be required for a wave speed of 25.00 m/southward?

Ii strings are attached between 2 poles separated by a distance of 2.00 m as shown below, both nether the same tension of 600.00 N. String 1 has a linear density of

${\mu }_{1}=0.0025\,\text{kg/m}$

and string 2 has a linear mass density of

${\mu }_{2}=0.0035\,\text{kg/m}.$

Transverse wave pulses are generated simultaneously at reverse ends of the strings. How much time passes earlier the pulses pass i another?
Figure shows two strings attached between two poles. A wave propagates from left to right in the top string with velocity v subscript w1. A wave propagates from right to left in the bottom string with velocity v subscript w2.

[reveal-answer q="431633″]Show Solution[/reveal-answer]
[hidden-answer a="431633″]

${v}_{1}t+{v}_{2}t=2.00\,\text{m},\,t=1.69\,\text{ms}$

[/hidden-answer]

Ii strings are attached between ii poles separated by a distance of 2.00 meters as shown in the preceding figure, both strings accept a linear density of

${\mu }_{1}=0.0025\,\text{kg/m},$

the tension in cord one is 600.00 N and the tension in cord 2 is 700.00 North. Transverse moving ridge pulses are generated simultaneously at opposite ends of the strings. How much fourth dimension passes before the pulses laissez passer one another?

The note

${E}_{4}$

is played on a piano and has a frequency of

$f=393.88.$

If the linear mass density of this string of the pianoforte is

$\mu =0.012\,\text{kg/m}$

and the string is nether a tension of chiliad.00 N, what is the speed of the moving ridge on the string and the wavelength of the wave?

[reveal-answer q="fs-id1165037159751″]Show Solution[/reveal-reply]

[hidden-answer a="fs-id1165037159751″]

$v=288.68\,\text{m/s},\,\lambda =0.73\,\text{m}$

[/hidden-answer]

2 transverse waves travel through a taut string. The speed of each wave is

$v=30.00\,\text{m/s}.$

A plot of the vertical position as a function of the horizontal position is shown below for the time

$t=0.00\,\text{s}.$

(a) What is the wavelength of each wave? (b) What is the frequency of each wave? (c) What is the maximum vertical speed of each string?
Two transverse waves are shown on a graph. The first one is labeled y1 parentheses x, t. Its y value varies from -3 m to 3 m. It has crests at x equal to 5 m and 15 m. The second wave is labeled y2 parentheses x, t. Its y value varies from -2 to 2. It has crests at x equal to 3 m, 9 m and 15 m.