CIE AS/A Level Physics 7.1 Progressive waves Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 7.1 Progressive waves Study Notes – New Syllabus
CIE AS/A Level Physics 7.1 Progressive waves Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- describe what is meant by wave motion as illustrated by vibration in ropes, springs and ripple tanks
- understand and use the terms displacement, amplitude, phase difference, period, frequency, wavelength and speed
- understand the use of the time-base and y-gain of a cathode-ray oscilloscope (CRO) to determine frequency and amplitude
- derive, using the definitions of speed, frequency and wavelength, the wave equation v = f λ
- recall and use v = f λ
- understand that energy is transferred by a progressive wave
- recall and use intensity = power/area and intensity ∝ (amplitude)2 for a progressive wave
Definition and Nature of Wave Motion
Wave motion is the transfer of energy from one point to another through vibrations or oscillations, without any net transfer of matter.
- In wave motion, particles of the medium oscillate about their mean positions.
- Energy is transmitted through the medium as a disturbance.
- Although energy travels, the particles themselves only vibrate — they do not move along with the wave.
Energy Transfer Example:
When you flick one end of a rope, the disturbance travels along the rope — energy moves along, but each part of the rope only moves up and down (it does not travel with the wave).
Types of Wave Motion
Depending on the direction of vibration of the particles relative to the direction of wave travel, waves are classified as:
- Transverse waves: The particles vibrate perpendicular to the direction in which the wave travels. Example: waves on a rope, light waves, water ripples.
- Longitudinal waves: The particles vibrate parallel to the direction of wave travel. Example: sound waves in air, compression waves in a spring.
Illustrations of Wave Motion
(a) Vibration in a Rope or String
- When one end of a rope is shaken up and down, a disturbance travels along its length.
- Each particle of the rope oscillates vertically, while the wave moves horizontally.
- This shows transverse wave motion.
- Energy is transferred along the rope; no section of the rope travels with the wave.
(b) Vibration in a Spring (Slinky)
- If the spring is pushed and pulled along its length, compressions and rarefactions are produced.
- Each coil vibrates back and forth (parallel to wave direction).
- This demonstrates longitudinal wave motion.
- Energy moves along the spring, while the coils oscillate about fixed positions.
(c) Ripple Tank
- In a ripple tank, waves are generated on the surface of water by a vibrating paddle.
- The water particles move in small circular paths — they rise and fall but do not travel with the wave.
- The disturbance spreads out as circular ripples, showing wave propagation and energy transfer.
- Ripples on water are examples of surface waves (combination of transverse and longitudinal motion).
Example
When a pulse travels along a stretched rope, what moves along the rope — the rope material or the energy? Explain how this shows the nature of wave motion.
▶️ Answer / Explanation
Only energy travels along the rope; the particles of the rope merely move up and down around their equilibrium positions. This demonstrates that in wave motion, energy is transmitted through the medium without the transport of matter.
Example
In an experiment, a student generates waves using
(i) a rope moved up and down, and
(ii) a spring compressed and released along its length. Compare the motion of the medium in each case and identify the type of wave.
▶️ Answer / Explanation
(i) In the rope, particles move perpendicular to the direction of the wave — a transverse wave is produced.
(ii) In the spring, particles move parallel to the direction of the wave — a longitudinal wave is produced.
Both cases show that wave motion transfers energy through vibration of particles, not mass transport.
Understanding Wave Quantities
In describing wave motion, several physical quantities are used to characterize how waves oscillate and propagate. These include displacement, amplitude, phase difference, period, frequency, wavelength, and wave speed.
Displacement \(( \mathrm{x} )\)
The displacement of a vibrating particle is its instantaneous distance and direction from its equilibrium (rest) position.
- It varies continuously with time and position.
- Displacement can be positive or negative, depending on the direction of vibration.
Units: metre (m)
Example: In a vibrating string, the displacement of a point is how far it moves above or below the equilibrium line at any moment.
Amplitude \(( \mathrm{A} )\)
The amplitude is the maximum displacement of a vibrating particle from its equilibrium position.
- It represents the “height” of the wave crest (or depth of the trough) from the mean position.
- It indicates the energy of the wave larger amplitude → higher energy.
Units: metre (m)
Example
A wave has a crest 0.03 m above and a trough 0.03 m below the equilibrium position. What is its amplitude?
▶️ Answer / Explanation
The amplitude is the maximum displacement from the mean position, not crest-to-trough distance. Hence \( \mathrm{A = 0.03\,m.} \)
Phase and Phase Difference \(( \mathrm{\phi} )\)
Phase:
The phase of a wave describes the stage of vibration a particle is in at a given time i.e., how far through a complete cycle it has moved.
- Phase is often expressed as an angle (in radians or degrees) or fraction of a period.
- A phase of \( \mathrm{0^\circ} \) and \( \mathrm{360^\circ} \) (or \( \mathrm{0} \) and \( \mathrm{2\pi} \) radians) represents the same point in a cycle.
Phase Difference:
The phase difference between two points on a wave is the fraction of a complete cycle by which one point’s vibration leads or lags behind the other.
\( \mathrm{\text{Phase difference} = \dfrac{\Delta x}{\lambda} \times 360^\circ \text{ (or } 2\pi \text{ radians)}} \)
- If two points vibrate in the same direction at the same time → they are in phase (phase difference = 0° or 360°).
- If two points move in opposite directions → they are out of phase (phase difference = 180°).
Example
Two points on a wave are separated by \( \mathrm{\dfrac{\lambda}{4}}. \) What is the phase difference between them?
▶️ Answer / Explanation
\( \mathrm{\text{Phase difference} = \dfrac{\Delta x}{\lambda} \times 360^\circ = \dfrac{1}{4} \times 360^\circ = 90^\circ.} \)
The two points are 90° (or \( \mathrm{\dfrac{\pi}{2}} \) radians) out of phase.
Period \(( \mathrm{T} )\)
The period is the time taken for one complete vibration or oscillation of a wave particle.
\( \mathrm{T = \dfrac{1}{f}} \)
- All particles on the same wave have the same period of vibration.
- Measured in seconds (s).
Frequency \(( \mathrm{f} )\)
The frequency of a wave is the number of complete oscillations (cycles) made per second.
\( \mathrm{f = \dfrac{1}{T}} \)
- Measured in hertz (Hz).
- 1 Hz = 1 cycle per second.
Relationship:
High frequency → short period; low frequency → long period.
Example
A sound wave has a frequency of \( \mathrm{250\,Hz.} \) What is its period?
▶️ Answer / Explanation
\( \mathrm{T = \dfrac{1}{f} = \dfrac{1}{250} = 0.004\,s = 4.0\,ms.} \)
Hence, the period of vibration is \( \mathrm{4.0\,ms.} \)
Wavelength \(( \mathrm{\lambda} )\)
The wavelength is the distance between two successive points on a wave that are in phase — for example, from crest to crest or from compression to compression.
- Measured along the direction of propagation.
- Unit: metre (m).
- Wavelength determines the spatial periodicity of the wave.
Wave Speed \(( \mathrm{v} )\)
The wave speed is the distance the wave travels per unit time — or the speed at which the wave disturbance moves through the medium.
\( \mathrm{v = f \lambda = \dfrac{\lambda}{T}} \)
- \( \mathrm{v} \): wave speed (m/s)
- \( \mathrm{f} \): frequency (Hz)
- \( \mathrm{\lambda} \): wavelength (m)
Example
A water wave has a frequency of \( \mathrm{5.0\,Hz} \) and a wavelength of \( \mathrm{0.80\,m.} \) Calculate the speed of the wave.
▶️ Answer / Explanation
\( \mathrm{v = f\lambda = 5.0 \times 0.80 = 4.0\,m/s.} \)
The speed of the wave is \( \mathrm{4.0\,m/s.} \)
The Cathode-Ray Oscilloscope (CRO)
The cathode-ray oscilloscope (CRO) is an instrument used to display and measure alternating voltage signals as a function of time. It allows the user to determine the frequency and amplitude of a waveform accurately.
Main Controls
- Time-base (horizontal, x-axis): Controls how fast the spot moves horizontally across the screen i.e. how much time each division represents. It determines the time scale of the waveform display.
- Y-gain (vertical, y-axis): Controls the vertical deflection per unit voltage i.e. how many volts correspond to one vertical division on the screen. It determines the voltage (or amplitude) scale of the display.
Determining Amplitude from CRO
Amplitude is found from the peak vertical deflection of the waveform on the screen.
\( \mathrm{V_{peak} = (\text{number of vertical divisions}) \times (\text{Y-gain})} \)
- Y-gain gives volts per division (V/div).
- The vertical distance from the centre line to the waveform peak represents the amplitude.
- For peak-to-peak value: multiply amplitude by 2.
Determining Frequency from CRO
The time-base setting gives the time represented by one horizontal division.
\( \mathrm{T = (\text{number of horizontal divisions for one cycle}) \times (\text{time-base setting})} \)
Once the period \( \mathrm{T} \) is known, the frequency is calculated from:
\( \mathrm{f = \dfrac{1}{T}} \)
Example
On a CRO screen, an alternating voltage waveform covers 4 vertical divisions from peak to peak and 5 horizontal divisions for one full cycle. The Y-gain is set to \( \mathrm{2.0\,V/div} \) and the time-base to \( \mathrm{0.5\,ms/div.} \) Calculate the amplitude and frequency of the signal.
▶️ Answer / Explanation
Step 1: Amplitude
The waveform covers 4 vertical divisions peak-to-peak → amplitude is half of that:
\( \mathrm{V_{peak} = \dfrac{4}{2} \times 2.0 = 4.0\,V.} \)
Step 2: Frequency
One cycle = 5 divisions; time-base = \( \mathrm{0.5\,ms/div.} \)
\( \mathrm{T = 5 \times 0.5\,ms = 2.5\,ms = 2.5\times10^{-3}\,s.} \)
\( \mathrm{f = \dfrac{1}{T} = \dfrac{1}{2.5\times10^{-3}} = 400\,Hz.} \)
Result: Amplitude = \( \mathrm{4.0\,V,} \) Frequency = \( \mathrm{400\,Hz.} \)
Derivation of the Wave Equation \(\mathrm{v = f\lambda}\)
Fundamental Idea
The wave equation \( \mathrm{v = f\lambda} \) connects the speed of a wave (\( \mathrm{v} \)) with its frequency (\( \mathrm{f} \)) and wavelength (\( \mathrm{\lambda} \)). It expresses how far a wave travels in one second based on how often each cycle repeats and the length of each cycle.
Definitions Used in the Derivation
- Wave speed (\( \mathrm{v} \)\): the distance the wave travels per unit time (m/s).
- Wavelength (\( \mathrm{\lambda} \)): the distance between two successive points that are in phase — e.g., crest to crest (m).
- Frequency (\( \mathrm{f} \)): the number of complete oscillations (or cycles) produced per second (Hz).
- Period (\( \mathrm{T} \)): the time for one complete oscillation (s), related by \( \mathrm{f = \dfrac{1}{T}}. \)
Step-by-Step Derivation
Step 1: Define wave speed
The speed of a wave is the distance travelled by the wave in one second:
\( \mathrm{v = \dfrac{\text{distance travelled}}{\text{time taken}}.} \)
Step 2: Express the distance travelled in terms of wavelength
In one complete oscillation, the wave travels forward by one wavelength (\( \mathrm{\lambda} \)). The time taken for one oscillation is the period (\( \mathrm{T} \)).
\( \mathrm{v = \dfrac{\lambda}{T}.} \)
Step 3: Substitute the relationship ( \mathrm{f = \dfrac{1}{T}} )
Replacing \( \mathrm{T} \) in the above equation gives:
\( \mathrm{v = \lambda \times f.} \)
4. Final Expression
\( \mathrm{v = f\lambda} \)
This is the wave equation, relating speed, frequency, and wavelength.
5. Physical Meaning
- As frequency increases (more cycles per second), wave speed increases if wavelength remains constant.
- If the medium is constant, wave speed \( \mathrm{v} \) is fixed — so increasing frequency means the wavelength must shorten.
- The relationship applies to all types of waves — mechanical (e.g. sound, water) and electromagnetic (e.g. light, radio).
Example
A water wave passes a stationary point 20 times every 10 seconds. The distance between two successive crests is \( \mathrm{0.60\,m.} \) Calculate
(a) the frequency, and
(b) the speed of the wave.
▶️ Answer / Explanation
(a) Frequency:
\( \mathrm{f = \dfrac{\text{number of waves}}{\text{time}} = \dfrac{20}{10} = 2.0\,Hz.} \)
(b) Speed:
\( \mathrm{v = f\lambda = 2.0 \times 0.60 = 1.2\,m/s.} \)
Result: The wave travels at \( \mathrm{1.2\,m/s.} \)
Applying the Wave Equation \(\mathrm{v = f\lambda}\)
The Wave Equation
The fundamental relationship between wave speed, frequency, and wavelength is given by:
\( \mathrm{v = f\lambda} \)
- \( \mathrm{v} \): wave speed (m/s)
- \( \mathrm{f} \): frequency (Hz)
- \( \mathrm{\lambda} \): wavelength (m)
Meaning of Each Quantity
- Wave speed (v): how fast the disturbance travels through the medium.
- Frequency (f): how many complete cycles pass a fixed point each second.
- Wavelength (λ): the distance between corresponding points (e.g., crest to crest) on successive waves.
Rearranging the Equation
This formula can be rearranged depending on which variable is required:
| To find | Use |
|---|---|
| Wave speed | \( \mathrm{v = f\lambda} \) |
| Frequency | \( \mathrm{f = \dfrac{v}{\lambda}} \) |
| Wavelength | \( \mathrm{\lambda = \dfrac{v}{f}} \) |
Using the Wave Equation
The equation applies to all wave types, including:
- Sound waves: \( \mathrm{v} \) depends on the medium (e.g., \( \mathrm{343\,m/s} \) in air at 20°C).
- Water waves: \( \mathrm{v} \) depends on depth and wavelength.
- Light and electromagnetic waves: \( \mathrm{v = 3.0\times10^8\,m/s} \) in vacuum.
Example
Sound of frequency \( \mathrm{256\,Hz} \) has a wavelength of \( \mathrm{1.35\,m.} \) Calculate the speed of sound in air.
▶️ Answer / Explanation
Using \( \mathrm{v = f\lambda} \):
\( \mathrm{v = 256 \times 1.35 = 346\,m/s.} \)
The speed of sound in air ≈ \( \mathrm{3.46\times10^2\,m/s.} \)
Example
A radio station transmits at a frequency of \( \mathrm{100\,MHz} \) (i.e., \( \mathrm{1.00\times10^8\,Hz} \)). Calculate the wavelength of the radio wave in air.
▶️ Answer / Explanation
For electromagnetic waves in air: \( \mathrm{v = 3.0\times10^8\,m/s.} \)
\( \mathrm{\lambda = \dfrac{v}{f} = \dfrac{3.0\times10^8}{1.0\times10^8} = 3.0\,m.} \)
The wavelength of the radio wave is \( \mathrm{3.0\,m.} \)
Example
Water waves travel at \( \mathrm{2.5\,m/s} \) with a wavelength of \( \mathrm{0.50\,m.} \) Determine their frequency.
▶️ Answer / Explanation
Using \( \mathrm{f = \dfrac{v}{\lambda}} \):
\( \mathrm{f = \dfrac{2.5}{0.50} = 5.0\,Hz.} \)
The frequency of the water waves is \( \mathrm{5.0\,Hz.} \)
Graphical Connection
The wave equation can also be visualized on displacement–time or displacement–distance graphs:
- The horizontal spacing between crests on a displacement–distance graph = \( \mathrm{\lambda.} \)
- The time for one complete oscillation on a displacement–time graph = \( \mathrm{T.} \)
- Using \( \mathrm{v = \dfrac{\lambda}{T}} \) or \( \mathrm{v = f\lambda} \), wave speed connects both space and time representations.
Typical Values of Wave Speeds
| Type of Wave | Medium | Speed (m/s) |
|---|---|---|
| Sound | Air (20 °C) | 343 |
| Sound | Water | 1500 |
| Light | Vacuum | \( \mathrm{3.0\times10^8} \) |
| Water waves | Surface | 0.1 – 10 |
Key Takeaways
- The wave equation \( \mathrm{v = f\lambda} \) applies to all progressive waves.
- If frequency increases while speed remains constant, wavelength decreases.
- For electromagnetic waves in a vacuum, \( \mathrm{v = 3.0\times10^8\,m/s.} \)
- For mechanical waves, \( \mathrm{v} \) depends on the medium’s properties (e.g., tension, density).
Example
Light of wavelength \( \mathrm{600\,nm} \) passes from air into glass, where its speed becomes \( \mathrm{2.0\times10^8\,m/s.} \) Find
(a) its frequency, and
(b) its wavelength in glass.
▶️ Answer / Explanation
(a) Frequency (same in both media):
\( \mathrm{f = \dfrac{v}{\lambda} = \dfrac{3.0\times10^8}{600\times10^{-9}} = 5.0\times10^{14}\,Hz.} \)
(b) Wavelength in glass:
\( \mathrm{\lambda’ = \dfrac{v’}{f} = \dfrac{2.0\times10^8}{5.0\times10^{14}} = 4.0\times10^{-7}\,m = 400\,nm.} \)
Result: Frequency = \( \mathrm{5.0\times10^{14}\,Hz} \) Wavelength in glass = \( \mathrm{400\,nm.} \)
Energy Transfer by a Progressive Wave
A progressive wave is a wave that transfers energy from one point to another through the vibration of particles in a medium (or through an electromagnetic field), without transferring matter.
Nature of Energy Transfer
- As the wave travels, each particle (or field point) oscillates about its equilibrium position.
- These oscillations carry energy forward to neighbouring particles, producing a continuous transfer through the medium.
- The particles themselves do not move with the wave — they simply pass on the disturbance.
Examples of Energy Transfer in Waves
- Sound waves: Transfer kinetic and potential energy through air particles’ vibrations.
- Water waves: Transfer energy across the water surface; floating objects move up and down but not sideways.
- Light waves: Transfer electromagnetic energy through space — no medium required.
Example
When a wave travels along a stretched string, what is transferred from one end to the other: the string material, or energy? Explain briefly.
▶️ Answer / Explanation
Energy is transferred along the string, not the material itself.
Each section of the string oscillates up and down and passes on its energy to neighbouring sections, while remaining in roughly the same position overall. This is characteristic of a progressive wave.
Intensity of a Progressive Wave
The intensity of a wave is the rate of energy transfer per unit area perpendicular to the direction of wave propagation.
\( \mathrm{I = \dfrac{P}{A}} \)
- \( \mathrm{I} \): intensity (W·m⁻²)
- \( \mathrm{P} \): power transmitted by the wave (W)
- \( \mathrm{A} \): cross-sectional area normal to the wave direction (m²)
Relationship Between Intensity and Amplitude
For a progressive wave, the energy carried by the wave is proportional to the square of its amplitude.
\( \mathrm{I \propto A^2} \)
- If amplitude doubles, intensity increases by a factor of four.
- This applies to both mechanical and electromagnetic waves.
Physical meaning:
- Amplitude reflects how much energy each particle has during vibration.
- Therefore, larger amplitude → greater energy → greater power transmitted per unit area.
Factors Affecting Intensity
- Distance from source: As a wave spreads out, its energy is distributed over a larger area, so intensity decreases.
- Amplitude: Higher amplitude → higher energy → higher intensity.
- Medium properties: Absorption or damping in the medium reduces transmitted intensity.
Example
A loudspeaker emits a sound wave with a power output of \( \mathrm{2.0\,W.} \) The sound spreads out uniformly over a spherical surface of radius \( \mathrm{4.0\,m.} \) Calculate the intensity of the sound at this distance.
▶️ Answer / Explanation
For a spherical wave, area = \( \mathrm{4\pi r^2.} \)
\( \mathrm{A = 4\pi (4.0)^2 = 201\,m^2.} \)
Using \( \mathrm{I = \dfrac{P}{A}} \):
\( \mathrm{I = \dfrac{2.0}{201} = 9.95\times10^{-3}\,W·m^{-2}.} \)
The intensity of the sound wave = \( \mathrm{9.9\times10^{-3}\,W·m^{-2}.} \)
Example
Two waves of the same frequency travel through identical media. The amplitude of Wave B is twice that of Wave A. Compare their intensities.
▶️ Answer / Explanation
Since \( \mathrm{I \propto A^2,} \)
\( \mathrm{\dfrac{I_B}{I_A} = \left(\dfrac{A_B}{A_A}\right)^2 = (2)^2 = 4.} \)
The intensity of Wave B is four times greater than that of Wave A.