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AP Physics 2: Algebra Notes

6.1.3 Wave Speed in Media and Strings

AP Syllabus focus: 'Wave speed depends on the wave type and the medium properties. For a string, wave speed depends on tension and mass per length: v_string = sqrt(F_T/(m/L)).'

Wave speed is set by physical conditions, not chosen freely by the source. For AP Physics 2, the key idea is that different media and different string properties produce different propagation speeds.

Wave speed as a medium-dependent quantity

Wave speed describes how fast a disturbance moves through a material.

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A transverse wave travels to the right while the medium oscillates vertically, illustrating the distinction between propagation direction and particle motion. This kind of picture helps you interpret “disturbance” in the definition of wave speed as something that moves through the medium even when individual bits of the medium mainly oscillate in place. Source

A person can make small or large disturbances, but the medium determines how rapidly that disturbance is passed from one region to the next. On the same string, pulses with different shapes still travel with the same basic speed if the string properties remain unchanged.

Wave speed: The rate at which a wave disturbance propagates through a medium.

This is why wave speed is treated as a property of the system rather than a property of the source alone. Different wave types rely on different physical mechanisms. In some cases the medium’s elasticity is most important; in others, inertia or internal forces dominate. The specification’s phrase wave type and the medium properties means you should ask two questions: what kind of wave is it, and what features of the material control how easily the disturbance is transmitted?

Why medium properties matter

A wave moves because one part of a medium influences the next part. Two broad ideas determine how quickly that transfer happens:

  • Restoring effects tend to pull the medium back toward equilibrium.

  • Inertia resists changes in motion because the medium has mass.

A stronger restoring effect usually leads to a faster wave, because a disturbed region can pull or push neighboring regions more strongly. Greater inertia usually leads to a slower wave, because more mass must be accelerated before the disturbance can continue onward. These same ideas appear in many wave situations, but on a string they can be expressed especially clearly.

For a string, the restoring effect comes from the tension in the string.

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Standing-wave patterns for n=1,2,3n=1,2,3 on a string fixed at both ends, with nodes and antinodes labeled. While the shape (mode) can change, the wave speed on a given string is still set by the string’s tension and linear mass density; the mode mainly changes the allowed wavelengths and frequencies. Source

The inertia comes from how much mass is spread along its length.

Linear mass density: The mass per unit length of a string, written as m/Lm/L and often denoted by μ\mu.

A larger value of m/Lm/L means each meter of string contains more mass. That makes each meter harder to accelerate, so the wave speed is smaller. A smaller value of m/Lm/L means the string is lighter per unit length, so the disturbance can move along more quickly.

Wave speed on a string

For a uniform string, AP Physics 2 uses a specific relationship between speed, tension, and mass per length.

vstring=FTm/Lv_{string}=\sqrt{\dfrac{F_T}{m/L}}

vstringv_{string} = wave speed on the string, in meters per second

FTF_T = tension force in the string, in newtons

mm = mass of the string segment, in kilograms

LL = length of that string segment, in meters

This equation says that wave speed increases when tension increases and decreases when mass per length increases. It is often rewritten using m/L=μm/L=\mu, but the physics is identical. If the string is uniform, any convenient segment can be used to find m/Lm/L because the ratio is the same everywhere along the string.

Interpreting the equation

The square-root dependence matters. Wave speed does not change one-for-one with tension or linear density.

  • If FTF_T becomes 44 times larger, the speed becomes 22 times larger.

  • If FTF_T becomes 99 times larger, the speed becomes 33 times larger.

  • If m/Lm/L becomes 44 times larger, the speed becomes 1/21/2 as large.

  • If m/Lm/L becomes 1/41/4 as large, the speed becomes 22 times larger.

These proportional changes are often more useful than raw substitution because many AP questions ask you to compare situations rather than compute from scratch. A quick unit check also supports the formula: dividing newtons by kilograms per meter gives units of square meters per square second, and taking the square root gives meters per second.

Physical meaning of tension and mass per length

A wave on a string propagates because neighboring parts of the string pull on one another. If the string is pulled tighter, a displaced section exerts a stronger force on adjacent sections. That stronger interaction passes the disturbance forward more quickly. In contrast, if the string is heavy per unit length, each small section is more difficult to speed up, so the disturbance advances more slowly.

This makes the formula physically sensible. High tension helps the wave move. High linear mass density resists the motion. The wave speed reflects the balance between these two effects, not just one of them by itself. A very slack string does not transmit a disturbance efficiently, while a tight, light string can carry a pulse much faster.

Reasoning skills for AP questions

When working with string-speed questions, identify which quantity is changing and which quantity is staying constant. This prevents many common errors.

  • Changing the way the string is disturbed does not set the speed by itself.

  • Increasing tension increases the wave speed, but only through a square-root relationship.

  • Increasing mass while keeping the same length increases m/Lm/L and reduces speed.

  • Increasing length while keeping the same mass decreases m/Lm/L and increases speed, if the tension is unchanged.

  • Changing the total length of a uniform string does not automatically change wave speed; only the resulting values of tension and mass per length matter.

It is also important to focus on mass per length, not just total mass. Two strings can have the same total mass but different wave speeds if that mass is distributed differently along their lengths. Likewise, two strings can have the same length but different wave speeds if one string is denser along its length or is held under a different tension.

In AP Physics 2, the central habit is to treat wave speed as a property set by the medium. For strings, that medium property is captured by the combination of tension and linear mass density in the equation above.

FAQ

A wave travels by affecting nearby parts of the string one section at a time.

What matters locally is how much mass each small section contains, not how much mass exists somewhere else on the string. That is why $m/L$ is the important quantity.

Two strings can have the same total mass but different wave speeds if one is long and thin while the other is short and thick. Their mass distributions are different, so their linear mass densities are different.

The formula works best when the string is modeled simply.

Typical assumptions are:

  • the string is uniform

  • the tension is approximately constant

  • the sideways disturbance is small

  • the string is flexible rather than stiff

  • energy loss is small enough to ignore for the speed model

If these assumptions fail badly, the actual wave speed may differ from the ideal expression.

Yes. In a real system, tightening a string usually increases the tension, but it can also make the string slightly longer.

If the string’s mass stays the same while its length increases, then $m/L$ decreases somewhat. That means two effects may occur together:

  • higher $F_T$, which tends to increase speed

  • lower $m/L$, which also tends to increase speed

This is one reason real strings can behave a little more subtly than the simplest ideal model suggests.

Not exactly.

If tension changes from place to place, the wave speed becomes a local property. One section may transmit the disturbance faster than another section.

In that case, a single number is only an approximation. A more careful description would treat the speed as varying along the string, depending on the local values of tension and mass per length.

The material alone does not determine the speed on a string.

Two strings made of the same substance can still differ in:

  • diameter

  • mass per length

  • applied tension

For example, a thicker string usually has a larger $m/L$ than a thinner string of the same material. Under the same tension, the thicker string generally carries waves more slowly because each meter has more mass to accelerate.

Practice Questions

A wave pulse travels on a string. The tension in the string is increased while the string’s mass per length stays the same. State how the wave speed changes, and explain your answer using the string-speed relationship.
[2 marks]

  • States that the wave speed increases. (1)

  • Explains that with constant m/Lm/L, the equation shows vstringv_{string} increases as FTF_T increases, specifically through a square-root dependence. (1)

Two uniform strings are held under the same tension of 4040 N.

  • String A has mass 0.0200.020 kg and length 2.02.0 m.

  • String B has mass 0.0800.080 kg and length 2.02.0 m.

(a) Which string has the greater wave speed? Explain qualitatively.
(b) Calculate the wave speed on each string.
(c) The tension in String B is changed until its wave speed matches that of String A. Determine the new tension required.
[6 marks]

  • (a) Identifies String A as having the greater wave speed. (1)

  • (a) Explains that String A has the smaller mass per length, so it has less inertia per meter. (1)

  • (b) Finds m/Lm/L for String A as 0.0100.010 kg/m. (1)

  • (b) Finds vA=40/0.01063v_A=\sqrt{40/0.010}\approx 63 m/s. (1)

  • (b) Finds m/Lm/L for String B as 0.0400.040 kg/m and vB=40/0.04032v_B=\sqrt{40/0.040}\approx 32 m/s. (1)

  • (c) Determines that String B needs 44 times the original tension, so the new tension is 160160 N. (1)

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