Wave Parameters

Spec mapping: OCR H556 Module 4.4 — Waves: wavelength, frequency, period, amplitude, phase difference; the wave equation $v = f\lambda$v=fλ; path difference vs phase difference. (Refer to the official OCR H556 specification document for exact wording.)

Having established what a wave is in the previous lesson, we now need the quantitative vocabulary to describe one. This lesson introduces the six core parameters of any progressive wave — displacement, amplitude, wavelength, frequency, period and phase difference — together with the two essential relationships between them:

$v = f\lambda \qquad T = \frac{1}{f}$v=fλT=f1​

These relationships will be used in every remaining lesson of the module. Mastery of them is non-negotiable for A-Level physics, and they are the foundation for everything that follows: refraction (where frequency stays fixed but wavelength and speed change together), polarisation (where amplitude becomes the projection onto a transmission axis), superposition (where phase difference governs whether crest meets crest or crest meets trough), and stationary waves (where two oppositely-travelling progressive waves of equal frequency interfere).

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Displacement

Displacement (x or y) is the distance of a particle of the medium from its equilibrium position at a given instant, measured in the appropriate direction (perpendicular to the direction of travel for a transverse wave, parallel for a longitudinal wave).

Displacement is a signed quantity: it can be positive or negative depending on which side of the equilibrium position the particle is. Over one complete oscillation, a particle's displacement passes through zero twice (as it crosses equilibrium) and reaches its maximum value twice (once positive, once negative).

Unit: metre (m), though for small oscillations you will often see it quoted in millimetres or micrometres.

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Amplitude

The amplitude (A) of a wave is the maximum displacement of a particle from its equilibrium position. It is the peak value of the signed displacement — a strictly positive number.

For a sinusoidal wave, every particle has the same amplitude (ignoring damping). The amplitude is a property of the wave, set by how much energy it carries, and does not change as the wave propagates through a non-absorbing medium.

Key fact: the energy (and therefore power) carried by a wave is proportional to A². Doubling the amplitude quadruples the energy carried per unit time.

Exam Tip: Be careful with units. If a transverse water wave has amplitude 5 cm, this is the distance from the mean level to the crest, not the crest-to-trough distance (which would be 2A = 10 cm).

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Wavelength

The wavelength $\lambda$λ of a wave is the distance between two successive points in phase — typically taken as the distance between two neighbouring crests, or two neighbouring troughs, or two neighbouring compressions.

Unit: metre (m). For visible light, wavelength is in the range $400$400 to $700$700 nm (nanometres); for radio waves it can be metres or kilometres; for X-rays it is around $10^{-10}$10−10 m.

If you plot displacement against position at a fixed instant of time, the wavelength is the horizontal distance between two successive peaks of the graph. Equivalently, $\lambda$λ is the distance the wave pattern travels in one period — a fact that will return when we derive the wave equation below.

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Frequency

The frequency (f) of a wave is the number of complete oscillations (or complete waves) passing a point per unit time.

Unit: hertz (Hz), where 1 Hz = 1 oscillation per second = 1 s⁻¹.

Frequency is set by the source of the wave and does not change when the wave passes into a different medium. (This is important when we study refraction in Lesson 5: when light enters glass, its speed and wavelength both decrease, but its frequency stays the same. Colour is determined by frequency, which is why colour does not change on refraction.)

Typical frequencies you will meet at A-Level:

Phenomenon	Typical frequency
Deep infrasound	< 20 Hz
Audible sound	20 Hz – 20 kHz
Ultrasound	> 20 kHz
Mains AC (UK)	50 Hz
FM radio	~100 MHz
Microwave ovens	2.45 GHz
Visible light	~430–750 THz
X-rays	~10¹⁸ Hz

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Period

The period $T$T of a wave is the time for one complete oscillation (the time taken for one full wave to pass a fixed point).

Unit: second (s).

Period and frequency are reciprocals of one another:

$T = \frac{1}{f} \qquad \text{equivalently} \qquad f = \frac{1}{T}$T=f1​equivalentlyf=T1​

This is the simplest and most often-used wave relationship. If a speaker emits a pure $440$440 Hz tone (middle A on a piano), the period of the sound wave is $T = 1/440 = 2.27 \times 10^{-3}$T=1/440=2.27×10−3 s = $2.27$2.27 ms.

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The Wave Equation: $v = f\lambda$v=fλ

The single most important equation in wave physics is:

$v = f\lambda$v=fλ

where $v$v is the wave speed (m s$^{-1}$−1), $f$f is the frequency (Hz), and $\lambda$λ is the wavelength (m).

Where Does It Come From?

In one complete oscillation (one period $T$T), the wave advances by exactly one wavelength $\lambda$λ. Hence:

$v = \frac{\text{distance}}{\text{time}} = \frac{\lambda}{T}$v=timedistance​=Tλ​

Combining with $f = 1/T$f=1/T gives $v = f\lambda$v=fλ. This derivation is worth committing to memory — OCR sometimes asks you to derive the wave equation from first principles in a short-answer question.

What Determines Each Variable?

• $v$v (wave speed) is set entirely by the medium. For sound in air at $20$20 °C, $v \approx 340$v≈340 m s$^{-1}$−1; for light in vacuum, $v = c = 3.00 \times 10^8$v=c=3.00×108 m s$^{-1}$−1.
• $f$f (frequency) is set by the source and does not change as the wave moves between media.
• $\lambda$λ (wavelength) is whatever value is needed to satisfy $v = f\lambda$v=fλ given the current medium's speed and the source's frequency.

So when a wave passes from one medium to another:

• Frequency stays the same.
• Speed changes (different medium).
• Wavelength changes by the same ratio as the speed.

This fact is at the heart of refraction (covered later in this module).

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Worked Example 1 — Sound Wave

Q. A tuning fork emits sound at $512$512 Hz. The speed of sound in air is $340$340 m s$^{-1}$−1. Calculate the wavelength.

A. Rearranging $v = f\lambda$v=fλ:

$\lambda = \frac{v}{f} = \frac{340}{512} = 0.664 \text{ m}$λ=fv​=512340​=0.664 m

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Worked Example 2 — Light in Glass

Q. A light ray of frequency $5.0 \times 10^{14}$5.0×1014 Hz enters a block of glass in which the speed of light is $2.0 \times 10^8$2.0×108 m s$^{-1}$−1. Calculate (a) the wavelength in vacuum, (b) the wavelength in the glass.

A.

(a) In vacuum, $v = c = 3.0 \times 10^8$v=c=3.0×108 m s$^{-1}$−1:

$\lambda_0 = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7} \text{ m} = 600 \text{ nm}$λ0​=fc​=5.0×10143.0×108​=6.0×10−7 m=600 nm

(b) In glass, $v = 2.0 \times 10^8$v=2.0×108 m s$^{-1}$−1. Frequency is unchanged:

$\lambda_g = \frac{v}{f} = \frac{2.0 \times 10^8}{5.0 \times 10^{14}} = 4.0 \times 10^{-7} \text{ m} = 400 \text{ nm}$λg​=fv​=5.0×10142.0×108​=4.0×10−7 m=400 nm

The wavelength has reduced by a factor of $1.5$1.5 (the refractive index of the glass — foreshadowing the refraction lesson). Notice the colour is unchanged, because colour is set by frequency, not wavelength: light that is yellow in vacuum is still yellow in glass even though its wavelength is shorter.

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Worked Example 3 — Radio Wave

Q. A BBC radio station transmits on a wavelength of $1500$1500 m. Calculate (a) the frequency, (b) the period.

A.

(a) Radio waves are electromagnetic, so $v = c = 3.0 \times 10^8$v=c=3.0×108 m s$^{-1}$−1:

$f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{1500} = 2.0 \times 10^5 \text{ Hz} = 200 \text{ kHz}$f=λc​=15003.0×108​=2.0×105 Hz=200 kHz

(b) $T = 1/f = 1 / (2.0 \times 10^5) = 5.0 \times 10^{-6}$T=1/f=1/(2.0×105)=5.0×10−6 s = $5.0$5.0 μs.

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Phase and Phase Difference

Phase is a way of describing where in its oscillation cycle a particle currently is. It is measured as an angle — either in degrees (one complete cycle = $360°$360°) or in radians (one complete cycle = $2\pi$2π rad). The radian form is the natural one in physics because it slots directly into sinusoidal expressions such as $y = A\sin(\omega t)$y=Asin(ωt) where $\omega = 2\pi f$ω=2πf has units of rad s$^{-1}$−1.

Two particles on a wave are said to be in phase if they are at the same point in their oscillation cycles at the same time — for instance, both at maximum positive displacement, both moving in the same direction.

Two particles are in antiphase (or "$180°$180° out of phase", or "$\pi$π rad out of phase") if they are exactly half a cycle apart — one at maximum positive displacement, the other at maximum negative displacement, each moving opposite to the other.

The Phase Difference Formula

The phase difference $\phi$ϕ between two points on a progressive wave separated by distance $d$d (the path difference) is:

$\phi = \frac{2\pi d}{\lambda} \text{ (rad)} \qquad \phi = \frac{360° d}{\lambda} \text{ (deg)}$ϕ=λ2πd​ (rad)ϕ=λ360°d​ (deg)

So:

Path difference $d$d	Phase difference (rad)	Phase difference (deg)	State
$0$0	$0$0	$0°$0°	In phase
$\lambda/4$λ/4	$\pi/2$π/2	$90°$90°	Quarter cycle
$\lambda/2$λ/2	$\pi$π	$180°$180°	Antiphase
$3\lambda/4$3λ/4	$3\pi/2$3π/2	$270°$270°	Three-quarter cycle
$\lambda$λ	$2\pi$2π	$360°$360°	In phase again

The third row of this table — antiphase, $\pi$π rad, half a wavelength of path difference — is the destructive-interference condition that drives Young's double-slit and single-slit diffraction. The last row (path difference of exactly one wavelength brings the wave back into phase) is the foundation of constructive interference and the diffraction-grating equation $d\sin\theta = n\lambda$dsinθ=nλ. Hold onto it.

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Worked Example 4 — Phase Difference

Q. A wave of wavelength $0.25$0.25 m travels along a rope. Two points on the rope are $6.25$6.25 cm apart. Calculate the phase difference in (a) radians, (b) degrees.

A. $d = 6.25$d=6.25 cm $= 0.0625$=0.0625 m; $\lambda = 0.25$λ=0.25 m. Hence $d/\lambda = 0.25$d/λ=0.25, so:

(a) $\phi = 2\pi \times 0.25 = \pi/2$ϕ=2π×0.25=π/2 rad (a quarter cycle). (b) $\phi = 360° \times 0.25 = 90°$ϕ=360°×0.25=90°.

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Displacement-Time and Displacement-Position Graphs

Two different graphs are used to visualise waves, and it is vital to distinguish them:

1. Displacement vs time (for one fixed particle): the horizontal axis is time, and the horizontal distance between two successive peaks is the period $T$T. The amplitude is read off the vertical axis.
2. Displacement vs position (at one fixed instant): the horizontal axis is position along the direction of travel, and the horizontal distance between two successive peaks is the wavelength $\lambda$λ. The amplitude is again on the vertical axis.

Both graphs look like sinusoids, and they are easy to confuse. Always check what the horizontal axis represents. In OCR exam questions an explicit "displacement vs time at the point P" or "displacement vs position at the instant $t = 0$t=0" caption is the safest place to read which graph you are looking at.

A useful exercise: take a single travelling wave $y(x, t) = A\sin(kx - \omega t)$y(x,t)=Asin(kx−ωt) and freeze it at $t = 0$t=0 to get $y(x, 0) = A\sin(kx)$y(x,0)=Asin(kx), a sinusoid in $x$x with spatial period $2\pi/k = \lambda$2π/k=λ. Now fix $x = 0$x=0 instead: $y(0, t) = -A\sin(\omega t)$y(0,t)=−Asin(ωt), a sinusoid in $t$t with temporal period $2\pi/\omega = T$2π/ω=T. The same underlying wave produces both graphs; the question is which independent variable you choose to plot against.

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Angular Frequency $\omega$ω and Wavenumber $k$k

It is often convenient to convert frequency and wavelength to their "angular" counterparts, which slot directly into trigonometric arguments without spare factors of $2\pi$2π:

$\omega = 2\pi f = \frac{2\pi}{T} \qquad k = \frac{2\pi}{\lambda}$ω=2πf=T2π​k=λ2π​

Here $\omega$ω is the angular frequency (rad s$^{-1}$−1) and $k$k is the wavenumber (rad m$^{-1}$−1). Both are intensive properties of the wave that recur throughout undergraduate physics.

In terms of these, the wave equation $v = f\lambda$v=fλ becomes the elegant form:

$v = \frac{\omega}{k}$v=kω​

and the sinusoidal solution of the wave equation can be written compactly as:

$y(x, t) = A\sin(kx - \omega t)$y(x,t)=Asin(kx−ωt)

(or $y = A\cos(\omega t - kx)$y=Acos(ωt−kx), depending on the choice of phase reference). At A-Level the symbols $\omega$ω and $k$k are not on the formula sheet but they are worth knowing for the wider physics — circular motion, simple harmonic motion, quantum mechanics — that uses them.

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Worked Example 5 — From Frequency to Angular Frequency

Q. A simple-harmonic oscillator has frequency $f = 50$f=50 Hz. Calculate its angular frequency $\omega$ω and its period $T$T.

A. $\omega = 2\pi f = 2\pi \times 50 = 314$ω=2πf=2π×50=314 rad s$^{-1}$−1. $T = 1/f = 1/50 = 0.020$T=1/f=1/50=0.020 s = $20$20 ms. Note that $\omega T = 2\pi$ωT=2π exactly (one complete cycle). This is the always-true identity: angular frequency times period equals exactly $2\pi$2π radians — a single complete cycle.

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Relating Wavenumber to Wavelength