Part 2 · Chapter 16

Waves — I

From a flick of a rope to a guitar string singing — how a disturbance travels without carrying the medium along with it

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • The difference between transverse and longitudinal waves, and how a single snapshot is captured by the sinusoidal wave function \(y(x,t) = y_{m}\sin(kx - \omega t)\).
  • The language of waves — amplitude \(y_{m}\), angular wave number \(k = 2\pi/\lambda\), and angular frequency \(\omega = 2\pi/T = 2\pi f\).
  • Why the wave speed is set by the medium: \(v = \omega/k = \lambda f\), and on a stretched string \(v = \sqrt{\tau/\mu}\).
  • How a wave carries energy at average rate \(P_{\text{avg}} = \tfrac{1}{2}\mu v \omega^{2} y_{m}^{2}\), and that every wave obeys the wave equation \(\partial^{2}y/\partial x^{2} = (1/v^{2})\,\partial^{2}y/\partial t^{2}\).
  • The superposition principle: two waves combine to give amplitude \(2y_{m}\cos\tfrac{1}{2}\phi\) (interference), and oppositely traveling waves build standing waves with resonances \(f = nv/2L\).
Section 16-1

What Is Physics?

One of the great unifying ideas in physics is that the same mathematics describes a ripple on a pond, the note from a violin, a tremor running through the Earth, a radio signal, and the light reaching your eye. All of these are waves — disturbances that travel, carrying energy and information from one place to another without transporting matter along with them. Mechanical waves (water, sound, seismic) need a medium; electromagnetic and matter waves do not. This chapter builds the framework for mechanical transverse waves on a string, and the language we develop here will carry over, almost word for word, to every other wave in the book.

Section 16-2

Types of Waves

Flick the end of a stretched rope and a pulse runs along it: each bit of rope moves up and down — perpendicular to the direction the pulse travels. That is a transverse wave. Push and pull the end of a long spring (or speak into the air) and the elements oscillate back and forth along the travel direction; that is a longitudinal wave. In both cases nothing is permanently carried forward — the medium returns to where it started, and only the disturbance moves on.

A wave that repeats smoothly in space and time is described by a sinusoidal wave function. For a wave moving in the positive \(x\) direction, the transverse displacement of the string element at position \(x\) and time \(t\) is:

Sinusoidal traveling wave (moving in +x)
\[ y(x,t) = y_{m}\sin(kx - \omega t) \]
Here \(y_m\) is the amplitude (maximum displacement), \(k\) is the angular wave number, \(\omega\) is the angular frequency, and the whole argument \((kx - \omega t)\) is the phase. A wave moving in the −x direction is written \(y = y_m\sin(kx + \omega t)\) — the plus sign reverses the direction of travel.
The element does not travel — the pattern does. Fix your eye on one point of the string: it merely oscillates up and down with simple harmonic motion. Fix your eye instead on a crest: it glides steadily along the string at the wave speed. Distinguishing these two motions — the transverse motion of an element versus the longitudinal motion of the wave shape — is the single most important habit in this chapter.
Section 16-3

Wavelength and Frequency

Two quantities pin down the periodicity of a sinusoidal wave. The wavelength \(\lambda\) is the shortest distance (along \(x\)) over which the pattern repeats — the distance between successive crests in a snapshot. The period \(T\) is the shortest time over which the motion of one element repeats. From the requirement that the phase repeat every \(\lambda\) in space and every \(T\) in time:

Angular wave number and angular frequency
\[ k = \frac{2\pi}{\lambda} \qquad\qquad \omega = \frac{2\pi}{T} = 2\pi f \]
\(k\) (rad/m) counts radians of phase per metre; \(\omega\) (rad/s) counts radians of phase per second. The ordinary frequency \(f = 1/T\) (in hertz) is the number of full oscillations per second made by any one element.
Section 16-4

The Speed of a Traveling Wave

To follow a particular point on the wave — say a crest — we keep its phase constant. Setting \(kx - \omega t = \text{const}\) and differentiating shows the crest advances at a fixed speed: divide the distance one wavelength by the time one period.

Wave speed
\[ v = \frac{\omega}{k} = \frac{\lambda}{T} = \lambda f \]
The three forms are identical. A wave moving in +x has phase \((kx-\omega t)\); a wave moving in −x has phase \((kx+\omega t)\). Speed is always positive; the sign in the phase carries the direction.
Speed belongs to the medium, not the source. You might expect a louder (bigger amplitude) or higher-pitched (higher frequency) wave to travel faster, but it does not. The speed \(v\) is fixed by the properties of the medium alone. Change the frequency and the medium responds by changing the wavelength to keep \(\lambda f = v\) — the relation that ties the source to the medium.
Section 16-5

Wave Speed on a Stretched String

What, exactly, are the medium properties that set the speed on a string? Dimensional analysis (or a small-element force argument) points to just two: the tension \(\tau\), which provides the restoring force, and the linear density \(\mu = m/L\) (mass per unit length), which provides the inertia.

🎻
Wave speed on a stretched string
v = √(τ / μ)

Tighten the string (larger \(\tau\)) and waves travel faster; use a heavier string (larger \(\mu\)) and they travel slower. This is precisely why a guitar's thick bass strings sound lower than its thin treble strings, and why turning a tuning peg raises the pitch. Notice that neither frequency nor amplitude appears — the speed is a property of the string.

Section 16-6

Energy and Power of a Wave

As a wave passes, each string element oscillates, so it carries both kinetic energy (from its transverse motion) and elastic potential energy (from being stretched). Energy is fed in at the source and transported along the string. Averaging over a full cycle gives the average power — the average rate of energy transmission:

Average power transmitted by a sinusoidal wave on a string
\[ P_{\text{avg}} = \tfrac{1}{2}\,\mu v\,\omega^{2} y_{m}^{2} \]
Power scales with the squares of both the angular frequency and the amplitude. Doubling the amplitude quadruples the energy transport rate; the same is true for doubling the frequency.
Section 16-7

The Wave Equation

Apply Newton's second law to a small element of the string and you obtain a single differential equation that every wave traveling at speed \(v\) must satisfy. It links the curvature of the string (the second space derivative) to the transverse acceleration (the second time derivative):

The wave equation
\[ \frac{\partial^{2} y}{\partial x^{2}} = \frac{1}{v^{2}}\,\frac{\partial^{2} y}{\partial t^{2}} \]
Any function of the form \(y = h(kx \pm \omega t)\) is a solution, with \(v = \omega/k\). The sinusoid is just the most useful special case; pulses of any shape travel undistorted on an ideal string.
Section 16-8

Interference of Waves

What happens when two waves overlap? The principle of superposition says the net displacement is simply the algebraic sum of the individual displacements — waves pass through each other unchanged. For two identical waves traveling the same way but out of step by a phase constant \(\phi\), the sum is again a sinusoid, but with a new amplitude:

Interference of two equal waves with phase difference φ
\[ y'(x,t) = \Big[\,2y_{m}\cos\tfrac{1}{2}\phi\,\Big]\,\sin\!\big(kx - \omega t + \tfrac{1}{2}\phi\big) \]
The resultant amplitude is \(2y_m\cos\tfrac{1}{2}\phi\). With \(\phi = 0\) the waves are in step: fully constructive interference, amplitude \(2y_m\). With \(\phi = \pi\) (180°) they cancel: fully destructive interference, amplitude 0.
Phase difference is what matters. Two waves at full strength can add up to nothing, or to double, depending entirely on how their phases line up. A phase difference can come from the sources being out of step, or from the two waves traveling different path lengths to reach the same point — the idea behind every interference pattern you will meet later in optics.
Section 16-9

Phasors

When more than two waves combine, or when their amplitudes differ, trigonometric algebra gets clumsy. A phasor tames it: represent each wave as a rotating vector whose length is the wave's amplitude and whose angle is its phase. Add the phasors head to tail like ordinary vectors, and the length of the resultant phasor is the amplitude of the combined wave; its angle is the combined phase.

🧭
Phasor addition
amplitudes add as vectors, not as numbers

For two equal phasors separated by angle \(\phi\), vector addition reproduces the interference result \(y'_{m} = 2y_{m}\cos\tfrac{1}{2}\phi\). For unequal waves \(y_{m1}\) and \(y_{m2}\), the law of cosines gives \(y'_{m} = \sqrt{y_{m1}^{2} + y_{m2}^{2} + 2y_{m1}y_{m2}\cos\phi}\) — a single, reliable recipe.

Section 16-10

Standing Waves and Resonance

Now let two identical waves travel in opposite directions — as happens when a wave reflects off a fixed end. Their superposition no longer travels at all; it forms a standing wave in which the pattern stays put while the string oscillates in place:

Standing wave (two oppositely traveling waves)
\[ y'(x,t) = \big[\,2y_{m}\sin kx\,\big]\cos\omega t \]
The bracketed factor is a position-dependent amplitude. Nodes (always at rest) occur where \(\sin kx = 0\), i.e. \(x = n\lambda/2\). Antinodes (maximum swing) sit halfway between, where \(|\sin kx| = 1\).

A string clamped at both ends (length \(L\)) can only support standing waves that fit a whole number of half-wavelengths between the fixed ends — those are its resonances. Each allowed pattern is a harmonic:

🎶
Resonant frequencies of a string fixed at both ends
λ = 2L / n  ·  f = nv / 2L  (n = 1, 2, 3, …)

The lowest frequency (\(n = 1\)) is the fundamental or first harmonic; the rest are integer multiples of it. This discrete ladder of allowed notes is exactly what gives a plucked guitar string, an organ pipe, or a microwave cavity its characteristic pitch and tone.

Worked Examples

Putting It to Work

1 Reading a wave function

Problem. A transverse wave on a string is described (SI units) by \(y = 0.00327\,\sin(72.1x - 2.72t)\). Find its (a) amplitude, (b) wavelength, (c) period and frequency, and (d) speed and direction.

Solution. Match the function term by term to \(y = y_{m}\sin(kx - \omega t)\): read off \(y_{m}\), \(k\), and \(\omega\), then convert.

From k = 2π/λ, ω = 2π/T, v = ω/k
\[\begin{gathered} y_{m} = 3.27\,\mathrm{mm}, \qquad \lambda = \frac{2\pi}{k} = \frac{2\pi}{72.1} \approx 0.0871\,\mathrm{m} = 8.71\,\mathrm{cm} \\ T = \frac{2\pi}{\omega} = \frac{2\pi}{2.72} \approx 2.31\,\mathrm{s}, \qquad f = \frac{1}{T} \approx 0.433\,\mathrm{Hz} \\ v = \frac{\omega}{k} = \frac{2.72}{72.1} \approx 0.0377\,\mathrm{m/s}\ \ (+x\ \text{direction}) \end{gathered}\]

The minus sign in the phase tells us the wave moves in the +x direction at about 3.8 cm/s — slow, because both \(\omega\) and the period reflect a leisurely oscillation.

2 Wave speed on a stretched string

Problem. A string has linear density \(\mu = 5.00\,\mathrm{g/m}\) and is held under tension \(\tau = 80.0\,\mathrm{N}\). (a) Find the wave speed. (b) If a source drives it at \(f = 100\,\mathrm{Hz}\), what is the wavelength?

Solution. Use \(v = \sqrt{\tau/\mu}\) (convert \(\mu\) to kg/m), then \(\lambda = v/f\).

v = √(τ/μ), then λ = v/f
\[\begin{gathered} v = \sqrt{\frac{\tau}{\mu}} = \sqrt{\frac{80.0}{5.00 \times 10^{-3}}} = \sqrt{1.60 \times 10^{4}} \approx 126.5\,\mathrm{m/s} \\ \lambda = \frac{v}{f} = \frac{126.5}{100} \approx 1.27\,\mathrm{m} \end{gathered}\]

The speed depends only on the string; changing the driving frequency would change \(\lambda\), never \(v\).

3 Power carried by the wave

Problem. The string of Example 2 (\(\mu = 5.00\,\mathrm{g/m}\), \(v = 126.5\,\mathrm{m/s}\)) carries a 100 Hz wave of amplitude \(y_{m} = 5.00\,\mathrm{mm}\). What average power does it transmit?

Solution. Compute \(\omega = 2\pi f\), then apply \(P_{\text{avg}} = \tfrac{1}{2}\mu v\omega^{2}y_{m}^{2}\).

P_avg = ½ μ v ω² y_m²
\[\begin{gathered} \omega = 2\pi(100) \approx 628\,\mathrm{rad/s} \\ P_{\text{avg}} = \tfrac{1}{2}(5.00\times10^{-3})(126.5)(628)^{2}(5.00\times10^{-3})^{2} \approx 3.1\,\mathrm{W} \end{gathered}\]

Because power goes as \(\omega^{2}y_{m}^{2}\), halving the amplitude would cut the transmitted power to one-quarter.

4 Interference of two waves

Problem. Two identical waves of amplitude \(y_{m} = 9.8\,\mathrm{mm}\) travel the same direction on a string. Find the resultant amplitude if their phase difference is (a) \(\phi = 100^{\circ}\) and (b) \(\phi = 0.50\,\mathrm{rad}\).

Solution. Apply \(y'_{m} = 2y_{m}\cos\tfrac{1}{2}\phi\), taking care to halve the phase before the cosine.

Resultant amplitude y′ₘ = 2yₘ cos(φ/2)
\[\begin{gathered} \text{(a)}\quad y'_{m} = 2(9.8)\cos\!\big(50^{\circ}\big) \approx 19.6 \times 0.643 \approx 12.6\,\mathrm{mm} \\ \text{(b)}\quad y'_{m} = 2(9.8)\cos\!\big(0.25\,\mathrm{rad}\big) \approx 19.6 \times 0.969 \approx 19.0\,\mathrm{mm} \end{gathered}\]

A small phase difference (b) leaves the waves nearly in step, so the amplitude is close to the constructive maximum of \(2y_{m} = 19.6\,\mathrm{mm}\).

5 Resonant frequencies of a string

Problem. A string fixed at both ends has length \(L = 1.20\,\mathrm{m}\) and carries waves at \(v = 126.5\,\mathrm{m/s}\). Find its three lowest resonant frequencies.

Solution. Resonance requires \(f_{n} = nv/2L\) for \(n = 1, 2, 3, \dots\)

f_n = n v / 2L
\[\begin{gathered} f_{1} = \frac{(1)(126.5)}{2(1.20)} \approx 52.7\,\mathrm{Hz} \quad (\text{fundamental}) \\ f_{2} = 2f_{1} \approx 105\,\mathrm{Hz}, \qquad f_{3} = 3f_{1} \approx 158\,\mathrm{Hz} \end{gathered}\]

The overtones are exact integer multiples of the fundamental — the hallmark of a string fixed at both ends, and the reason such a string produces a clear musical pitch.

Review

Chapter Summary

✓ Wave function

\(y = y_{m}\sin(kx - \omega t)\) for +x travel; use \((kx + \omega t)\) for −x. The phase carries the direction.

✓ Wave language

\(k = 2\pi/\lambda\), \(\omega = 2\pi/T = 2\pi f\), amplitude \(y_{m}\).

✓ Wave speed

\(v = \omega/k = \lambda f\). On a string, \(v = \sqrt{\tau/\mu}\) — set by the medium alone.

✓ Energy transport

\(P_{\text{avg}} = \tfrac{1}{2}\mu v\omega^{2}y_{m}^{2}\) — power grows as the square of both \(\omega\) and \(y_{m}\).

✓ Wave equation

\(\partial^{2}y/\partial x^{2} = (1/v^{2})\,\partial^{2}y/\partial t^{2}\) — obeyed by any pattern \(h(kx \pm \omega t)\).

✓ Interference

Superposition gives amplitude \(2y_{m}\cos\tfrac{1}{2}\phi\): constructive at \(\phi=0\), destructive at \(\phi=\pi\).

✓ Phasors

Add waves as rotating vectors; resultant \(y'_{m} = \sqrt{y_{m1}^{2}+y_{m2}^{2}+2y_{m1}y_{m2}\cos\phi}\).

✓ Standing waves

\(y' = [2y_{m}\sin kx]\cos\omega t\); string fixed both ends resonates at \(f = nv/2L\).

Practice

Problems

Most problems reduce to three relations: the wave-language identities (\(k = 2\pi/\lambda\), \(\omega = 2\pi f\)), the speed relation (\(v = \lambda f = \sqrt{\tau/\mu}\)), and — for combined waves — superposition (\(2y_{m}\cos\tfrac{1}{2}\phi\) and \(f = nv/2L\)). Keep amplitude in metres and watch radians-vs-degrees in every cosine.

  1. A wave is written \(y = 6.0\,\sin(0.020\pi x + 4.0\pi t)\) (cm, s). Find its (a) amplitude, (b) wavelength, (c) frequency, (d) speed, and (e) direction of travel.
  2. The equation of a transverse wave is \(y = 2.0\,\sin(20x - 600t)\) in SI units. What are its wavelength, frequency, and speed?
  3. A sinusoidal wave travels at 80 m/s along a string. If its frequency is 25 Hz, what is the distance between two points that differ in phase by \(\pi/3\,\mathrm{rad}\)?
  4. The fastest transverse wave on a steel wire is limited by the breaking stress. A string of linear density 0.25 g/cm is under 12 N of tension. Find the wave speed.
  5. What tension is required to make a 2.00 m, 60.0 g string carry transverse waves at 50.0 m/s?
  6. A 120 Hz wave of amplitude 1.6 mm travels on a string with \(\mu = 1.6\,\mathrm{g/m}\) under 90 N tension. At what average rate does it transmit energy?
  7. By what factor must the amplitude of a wave on a string be increased to double the average transmitted power, all else fixed?
  8. Two identical waves of amplitude 4.0 mm interfere on a string. Find the resultant amplitude when the phase difference is (a) 0, (b) \(\pi/2\), (c) \(2\pi/3\), and (d) \(\pi\).
  9. Three waves of equal amplitude 3.0 mm and phases 0, \(\pi/3\), and \(2\pi/3\) combine. Use phasors to find the resultant amplitude.
  10. A string fixed at both ends is 0.75 m long and supports waves at 240 m/s. Find (a) the fundamental frequency and (b) the frequency of the third harmonic.
  11. A standing wave on a 1.5 m string shows four antinodes when driven at 120 Hz. What is the wave speed?
  12. A nylon guitar string has \(\mu = 7.2\,\mathrm{g/m}\) and a vibrating length of 64 cm. What tension tunes its fundamental to 330 Hz?
Tip: always separate the two motions — a string element oscillates transversely with SHM (use \(u = \partial y/\partial t\) for its velocity), while the wave shape glides along at \(v = \sqrt{\tau/\mu}\). For interference, the phase difference is everything; for resonance on a string fixed at both ends, only the discrete frequencies \(f_{n} = nv/2L\) are allowed. When in doubt, sketch the snapshot and the standing-wave pattern before reaching for a formula.