(B) at What Instant Is There the Greatest Tension in the Cords? (How Do You Know?)

xvi Waves

sixteen.3 Wave Speed on a Stretched String

Learning Objectives

Past the stop of this section, you will be able to:

Determine the factors that affect the speed of a wave on a string
Write a mathematical expression for the speed of a wave on a cord and generalize these concepts for other media

The speed of a wave depends on the characteristics of the medium. For case, 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 take dissimilar thickness but may be made of similar material. They have dissimilar linear densities, where the linear density is divers as the mass per length,

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

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

$(\text{Δ}m)$

of a small-scale length of cord

$(\text{Δ}x)$

$\text{Δ}m=\mu \text{Δ}x.$

For example, if the string has a length of 2.00 chiliad 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 1.00-mm section is cut from the string, the mass of the i.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 too has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around 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 sound produced is proportional to the wave speed.

Moving ridge Speed on a String under Tension

To meet how the speed of a moving ridge on a string depends on the tension and the linear density, consider a pulse sent down a taut string ((Figure)). When the taut string is at rest at the equilibrium position, the tension in the cord

${F}_{T}$

is abiding. Consider a small element of the string with a mass equal to

$\text{Δ}m=\mu \text{Δ}x.$

The mass element is at residual and in equilibrium and the force of tension of either side of the mass chemical 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 16.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 element is equal in magnitude and reverse in direction.

If yous pluck a string under tension, a transverse wave moves in the positive ten-direction, as shown in (Figure). The mass element is minor only is enlarged in the figure to make it visible. The pocket-size mass element oscillates perpendicular to the wave motion as a result of the restoring force provided by the string and does not move in the ten-management. 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 16.14** A string under tension is plucked, causing a pulse to move along the string in the positive x-direction.

Assume that the inclination of the displaced cord with respect to the horizontal axis is small. The net forcefulness on the element of the string, interim parallel to the cord, is the sum of the tension in the string and the restoring force. The 10-components of the force of tension cancel, so the net force is equal to the sum of the y-components of the force. The magnitude of the x-component of the forcefulness is equal to the horizontal force 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 slope of a function at a point, which is equal to the fractional derivative of y with respect to x at that point. Therefore,

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

is equal to the negative gradient of the string at

${x}_{1}$

and

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

is equal to the slope 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 force is on the pocket-sized mass element can 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 second police force, the cyberspace force is equal to the mass times the acceleration. The linear density of the string

$\mu$

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

$\text{Δ}x$

approaches aught,

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

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, nosotros run into that the speed of the wave on a string depends on the tension and the linear density.

Speed of a Wave on a String Nether Tension

The speed of a pulse or wave on a string under tension tin can exist found 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.

Example

The Wave Speed of a Guitar Leap

On a six-string guitar, the high E string has a linear density of

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

and the low E cord has a linear density of

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

(a) If the high E cord is plucked, producing a moving ridge in the cord, what is the speed of the moving ridge if the tension of the cord is 56.xl N? (b) The linear density of the low E string is approximately 20 times greater than that of the high East string. For waves to travel through the low East cord at the same wave speed as the high Due east, would the tension demand to be larger or smaller than the loftier E string? What would be the judge tension? (c) Summate the tension of the depression Eastward cord needed for the same moving ridge speed.

Strategy

The speed of the wave can exist constitute 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 almost 20, the tension would need to be increased by a gene of xx.
Knowing the velocity and the linear density, the velocity equation can be solved for the force of tension
${F}_{T}=\mu {v}^{2}.$

Solution

Apply the velocity equation to notice 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 need to exist increased by a factor of approximately 20. The tension would be slightly less than 1128 N.

Use the velocity equation to find the actual 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 cord (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 cord and the wavelength of the waves. The six strings accept dissimilar linear densities and are "tuned" by changing the tensions in the strings. Nosotros will see in Interference of Waves that the wavelength depends on the length of the strings and the purlieus atmospheric condition. To play notes other than the primal notes, the lengths of the strings are inverse by pressing down on the strings.

Check Your Agreement

The wave speed of a wave on a cord 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?

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

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

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

$\sqrt{2}.$

[/hidden-answer]

Speed of Pinch Waves in a Fluid

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

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

The elastic property describes the trend of the particles of the medium to return 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 moving ridge through a liquid or gas depends on the density of the fluid and the bulk modulus of the fluid,

$v=\sqrt{\frac{Β}{\rho }}.$

Here the bulk modulus is divers 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 book, and

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

is the mass per unit volume. For example, sound is a mechanical moving ridge that travels through a fluid or a solid. The speed of sound in air with an atmospheric pressure level 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 detail 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 length of the cord.
In general, the speed of a moving ridge depends on the square root of the ratio of the elastic holding 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 past a factor of four, by what factor would the wave speed of a wave on the string increment?

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

Does a audio wave motion faster in seawater or fresh water, if both the ocean water and fresh h2o are at the aforementioned temperature and the sound 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 lowest density string and the highest density string are under the same tension, which string would support waves with the higher wave speed?

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

Shown beneath are three waves that were sent down a cord at different times. The tension in the string remains abiding. (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 connected by ii utility poles are sometimes heard to hum when driven into oscillation by 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″]Show Solution[/reveal-answer]
[subconscious-reply a="960058″]The tension in the wire is due to the weight of the electric power cable.[/hidden-reply]

Two strings, one with a low mass density and one with a high linear density are spliced together. The higher density stop is tied to a lab post and a educatee holds the free end of the low-mass density string. The student gives the string a flip and sends a pulse down the strings. If the tension is the aforementioned in both strings, does the pulse travel at the same wave velocity in both strings? If not, where does it travel faster, in the depression density cord or the high density string?

Issues

Transverse waves are sent along a 5.00-grand-long cord with a speed of 30.00 m/south. The string is under a tension of x.00 Northward. What is the mass of the cord?

A copper wire has a density of

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

a radius of i.20 mm, and a length L. 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-answer q="492763″]Bear witness Solution[/reveal-answer]
[hidden-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 string. The cord is plucked, sending a pulse down the string. Approximate the speed of the pulse as it moves downwards the cord.

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

[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 three meters. The cord is plucked and it takes 0.20 s for the pulse to reach the end of the cord. What is the tension of the string?

A cord is three.00 m long with a mass of 5.00 yard. The string is held taut with a tension of 500.00 North applied to the string. A pulse is sent down the string. How long does it accept the pulse to travel the iii.00 g of the string?

[reveal-answer q="439036″]Evidence Solution[/reveal-respond]
[subconscious-answer a="439036″]

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

[/hidden-answer]

Two strings are attached to poles, yet the first string is twice as long as the 2d. If both strings take the same tension and mu, what is the ratio of the speed of the pulse of the wave from the first string to the 2d string?

Two strings are fastened to poles, nonetheless the first string is twice the linear mass density mu of the second. If both strings accept the same tension, what is the ratio of the speed of the pulse of the moving ridge from the offset string to the 2nd string?

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

[hidden-reply a="fs-id1165037203500″]

0.707

[/hidden-reply]

Transverse waves travel through a string where the tension equals seven.00 N with a speed of 20.00 k/due south. What tension would exist required for a wave speed of 25.00 m/s?

Two strings are fastened between two poles separated by a distance of 2.00 m equally shown below, both under the aforementioned tension of 600.00 N. Cord i 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 moving ridge pulses are generated simultaneously at opposite ends of the strings. How much time passes before the pulses laissez passer one some other?
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-respond a="431633″]

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

[/hidden-answer]

Two strings are fastened between two poles separated by a distance of ii.00 meters as shown in the preceding figure, both strings have a linear density of

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

the tension in cord 1 is 600.00 Due north and the tension in string 2 is 700.00 N. Transverse wave pulses are generated simultaneously at opposite ends of the strings. How much fourth dimension passes earlier the pulses laissez passer one another?

The note

${E}_{4}$

is played on a pianoforte and has a frequency of

$f=393.88.$

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

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

and the cord is under a tension of 1000.00 N, what is the speed of the moving ridge on the cord and the wavelength of the wave?

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

[subconscious-respond a="fs-id1165037159751″]

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

[/hidden-reply]

Two transverse waves travel through a taut string. The speed of each moving ridge 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 fourth dimension

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

(a) What is the wavelength of each wave? (b) What is the frequency of each moving ridge? (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.