16-2 Wave Speed on a Stretched String#
Prompts
What determines the speed of a wave on a stretched string? Does it depend on the wave’s frequency or amplitude?
Define linear density \(\mu\). If a string has mass \(m\) and length \(L\), what is \(\mu\)?
Write the formula relating wave speed \(v\) to tension \(\tau\) and linear density \(\mu\). Why does a tighter string produce faster waves?
You increase the frequency of oscillations at one end of a string. Does the wave speed change? Does the wavelength change?
You increase the tension in the string. Does the wave speed change? Does the wavelength change?
Lecture Notes#
Overview#
The wave speed \(v\) on a stretched string is set by properties of the string—tension and mass per unit length—not by the wave’s frequency or amplitude.
Tighter string (larger tension) → faster waves. Heavier string (larger linear density) → slower waves.
From \(v = \lambda f\): the source fixes \(f\); the string fixes \(v\); therefore \(\lambda = v/f\) is determined.
Linear density and tension#
Linear density \(\mu\): mass per unit length. For a uniform string of mass \(m\) and length \(L\):
Tension \(\tau\): the magnitude of the force pulling the string taut at each end. A wave can only propagate if the string is under tension—adjacent elements pull on each other via the tension.
Wave speed formula#
The speed of a wave on a stretched string is
Dimensional argument: \(v\) has dimension LT\(^{-1}\); \(\tau\) (force) has MLT\(^{-2}\); \(\mu\) has ML\(^{-1}\). The only combination giving speed is \(\sqrt{\tau/\mu}\).
Physical argument: Tension provides the restoring force when the string is displaced; linear density provides the inertia. Higher \(\tau\) → stronger restoring force → faster response. Higher \(\mu\) → more inertia → slower response.
Important
Wave speed \(v\) depends only on \(\tau\) and \(\mu\). It does not depend on frequency \(f\), amplitude \(y_m\), or wavelength \(\lambda\).
Poll: Wave speed ratio
Two strings with different linear mass density are tied at the center and attached to walls with tension 1000 N. If \(\mu_1/\mu_2 = 25/9\), what is \(v_1/v_2\)?
(A) 9/25
(B) 3/5
(C) 25/9
(D) 5/3
(E) 1
Poll: Frequency at a knot
Same setup: two strings tied at the center. What is \(f_1/f_2\)?
(A) 9/25
(B) 3/5
(C) 25/9
(D) 5/3
(E) 1
Frequency, wavelength, and the source#
From section 16-1, \(v = \lambda f\). Rearranging:
The source (e.g., your hand oscillating the end) sets the frequency \(f\).
The string sets the wave speed \(v\).
The wavelength \(\lambda\) is then fixed by \(\lambda = v/f\).
Change |
Wave speed \(v\) |
Wavelength \(\lambda\) |
|---|---|---|
Increase \(f\) (source) |
Unchanged |
Decreases |
Increase \(\tau\) (tension) |
Increases |
Increases (if \(f\) fixed) |
Increase \(\mu\) (heavier string) |
Decreases |
Decreases (if \(f\) fixed) |
Poll: Wave appearance in two strings
Same two-string setup. Ignoring reflections, how do the waves compare in the two strings?
[FIGURE: Two strings tied at center; one with larger \(\mu\) (thicker) and one with smaller \(\mu\); optional snapshot showing different \(\lambda\)]
(A) Same wavelength and amplitude
(B) Same wavelength, different amplitude
(C) Different wavelength, same amplitude
(D) Different wavelength and amplitude
Poll: Frequency and tension
You send a wave along a string by oscillating one end. If you increase the frequency of oscillation, what happens to (a) the wave speed and (b) the wavelength?
(A) Both increase
(B) Speed unchanged, wavelength decreases
(C) Both decrease
(D) Speed increases, wavelength unchanged
Summary#
Linear density \(\mu = m/L\); tension \(\tau\) = force stretching the string.
Wave speed \(v = \sqrt{\tau/\mu}\)—set by the string, not by \(f\) or amplitude.
\(\lambda = v/f\)—source fixes \(f\); string fixes \(v\); \(\lambda\) follows.