understand that energy is transferred by a progressive wave

Progressive (Travelling) Waves – Cambridge AS & A Level Physics (9702) – Syllabus 7.1

1. Key definitions (quick reference)

Quantity	Symbol	Unit	Definition
Displacement	\(y\)	metre (m)	Instantaneous distance of a particle from its equilibrium position.
Amplitude	\(A\)	metre (m)	Maximum displacement from equilibrium.
Period	\(T\)	second (s)	Time for one complete cycle; \(T=1/f\).
Frequency	\(f\)	hertz (Hz)	Number of cycles per second; \(f=1/T\).
Angular frequency	\(\omega\)	rad s\(^{-1}\)	\(\omega = 2\pi f\).
Wavelength	\(\lambda\)	metre (m)	Distance between two successive points that are in phase (e.g., crest‑to‑crest).
Wave‑number	\(k\)	rad m\(^{-1}\)	\(k = 2\pi/\lambda\).
Phase difference	\(\phi\)	radian (rad)	Angular shift between two points on the same wave.
Wave speed	\(v\)	metre s\(^{-1}\)	Speed at which a given wave‑profile travels through the medium.

2. Physical description of a progressive wave

• A progressive (travelling) wave is a disturbance that moves through a medium, carrying **energy** from one region to another.
• Each particle of the medium oscillates about its equilibrium position; the particles themselves do **not** travel with the wave.
• The overall shape of the wave profile is unchanged – it is simply shifted in space as time advances.
• Because the medium’s particles only move locally, the wave can transport energy without transporting mass.

3. Mathematical description

For a sinusoidal wave travelling in the + x direction the transverse displacement is

\[ y(x,t)=A\cos\!\bigl(kx-\omega t+\phi\bigr) \]

Key points:

• The term \(-\omega t\) gives a wave moving in the + x direction; changing the sign to \(+\omega t\) gives a wave travelling in the – x direction.
• If the sign of the spatial term is reversed (i.e. \(kx+\omega t\)) while the time term remains \(-\omega t\), the result is a **standing wave** – a pattern that does not travel but oscillates in place. This distinction is required later in the syllabus.

4. Derivation of the wave‑speed relation \(v = f\lambda\)

1. For a point of constant phase the argument of the cosine is constant: \[ kx-\omega t+\phi = \text{constant}. \]
2. Differentiate with respect to time (phase held constant): \[ k\frac{dx}{dt}-\omega = 0\;\;\Longrightarrow\;\;\frac{dx}{dt}= \frac{\omega}{k}. \]
3. The quantity \(\dfrac{dx}{dt}\) is the speed \(v\) of the wave‑profile, so \[ v = \frac{\omega}{k}. \]
4. Substituting \(\omega = 2\pi f\) and \(k = 2\pi/\lambda\) gives the familiar form \[ v = f\lambda . \]

5. Energy in a transverse wave on a string

We consider a thin string under a uniform tension \(T\) and linear mass density \(\mu\). The analysis assumes small‑amplitude (linear) waves so that the tension remains essentially constant.

• Kinetic energy of a short element \(\Delta x\): \[ K = \frac{1}{2}\mu\left(\frac{\partial y}{\partial t}\right)^{2}\Delta x . \]
• Potential (elastic) energy stored by the extra stretch of the element: \[ U = \frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^{2}\Delta x . \]

The instantaneous energy per unit length is therefore

\[ \mathcal{E}(x,t)=\frac{1}{2}\mu\left(\frac{\partial y}{\partial t}\right)^{2} +\frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^{2}. \]

6. Average energy density, power and intensity

For a harmonic wave the time‑average (over one period) of the kinetic and potential parts are equal. Substituting the sinusoidal form \(y=A\cos(kx-\omega t)\) gives

\[ \langle K\rangle = \langle U\rangle = \frac{1}{4}\mu\omega^{2}A^{2}, \qquad \langle\mathcal{E}\rangle = \frac{1}{2}\mu\omega^{2}A^{2}. \]

Average power transmitted past a fixed point on the string is the amount of energy crossing that point per unit time. Because the wave profile moves with speed \(v\), the energy crossing per second is simply the product of the energy density and the speed:

\[ \boxed{\;\langle P\rangle = v\,\langle\mathcal{E}\rangle = \frac{1}{2}\mu v\omega^{2}A^{2}\;} \]

Intensity is defined as power per unit area. For a string the “area” is the cross‑sectional area of the string, which is effectively 1 m of length (so intensity has the same numerical value as power per metre). For waves that spread over a real surface (e.g. sound or light) the same definition applies:

\[ I = \frac{\langle P\rangle}{A_{\text{cross}}}. \]

Summary of key relationships

Quantity	Expression	Physical meaning
Wave speed	\(v = \dfrac{\omega}{k}=f\lambda\)	How fast the wave profile moves.
Average kinetic energy density	\(\langle K\rangle = \dfrac{1}{4}\mu\omega^{2}A^{2}\)	Mean kinetic energy per metre of string.
Average potential energy density	\(\langle U\rangle = \dfrac{1}{4}\mu\omega^{2}A^{2}\)	Mean elastic energy per metre.
Average total energy density	\(\langle\mathcal{E}\rangle = \dfrac{1}{2}\mu\omega^{2}A^{2}\)	Sum of kinetic and potential contributions.
Average power transmitted	\(\langle P\rangle = \dfrac{1}{2}\mu v\omega^{2}A^{2}\)	Rate at which energy crosses a point on the string.
Intensity (general)	\(I = \dfrac{\langle P\rangle}{A_{\text{cross}}}\)	Power per unit area (for a string, \(A_{\text{cross}}=1\) m).

7. Measuring a travelling wave with a Cathode‑Ray Oscilloscope (CRO)

The CRO provides a direct visual record of the time‑variation of the displacement (or a voltage proportional to it).

1. Setup: Attach a lightweight probe (e.g., a mirror‑mounted photodiode or magnetic pickup) to the string at a fixed position. Connect the probe to the vertical (y‑gain) input of the CRO.
2. Time‑base (horizontal) control: Adjust the time‑base so that one complete cycle occupies a convenient number of horizontal divisions. If the trace spans \(n_{\text{div}}\) divisions and each division corresponds to \(\Delta t\) s, then \[ T = n_{\text{div}}\;\Delta t,\qquad f = \frac{1}{T}. \]
3. Vertical gain: The vertical scale (V / division) together with the measured peak‑to‑peak voltage \(V_{\text{pp}}\) gives the amplitude of the displacement: \[ A = \frac{V_{\text{pp}}}{2\;(\text{V/div})}\times\frac{1}{S}, \] where \(S\) is the probe’s sensitivity (metres per volt).
4. Angular frequency: Once \(f\) is known, \(\omega = 2\pi f\).
5. Verification of \(v = f\lambda\):
   • Measure \(\lambda\) by marking successive crests on the string (or by using two probes a known distance apart).
   • Obtain an independent speed measurement, e.g. by sending a short pulse and timing its travel over a known distance.
   • Check that the product \(f\lambda\) agrees with the measured speed within experimental uncertainties.

8. Example calculation

Problem: A string under tension \(T = 50\;\text{N}\) has linear mass density \(\mu = 0.020\;\text{kg m}^{-1}\). It is driven at frequency \(f = 25\;\text{Hz}\) with amplitude \(A = 2.0\;\text{mm}\). Find the average power transmitted along the string.

Solution:

1. Wave speed from the tension‑mass relation: \[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{50}{0.020}} = 50\;\text{m s}^{-1}. \]
2. Angular frequency: \[ \omega = 2\pi f = 2\pi(25) = 157\;\text{rad s}^{-1}. \]
3. Amplitude in metres: \(A = 2.0\times10^{-3}\;\text{m}\).
4. Average power: \[ \langle P\rangle = \frac{1}{2}\mu v\omega^{2}A^{2} = \frac{1}{2}(0.020)(50)(157)^{2}(2.0\times10^{-3})^{2} \approx 0.25\;\text{W}. \]

The calculation illustrates that even a modest amplitude can transport a measurable amount of energy when the frequency (hence \(\omega\)) is relatively high.

9. Common misconceptions

• “The wave carries mass.” – Only the medium’s particles move locally; the wave transports **energy**, not mass.
• “Amplitude does not affect energy.” – Energy (and thus power) is proportional to \(A^{2}\); doubling the amplitude quadruples the transmitted power.
• “Higher frequency means each particle moves faster.” – The maximum particle speed is \(\omega A\); increasing frequency raises this speed, but the particle still only oscillates about its equilibrium position.
• “A standing wave also transports energy.” – In an ideal standing wave the net energy flow is zero; energy is stored locally in the alternating kinetic and potential forms.