\$v = f\lambda\$
where v is the speed, f the frequency and λ the wavelength.
\$\frac{\partial^{2}y}{\partial x^{2}}=\frac{1}{v^{2}}\frac{\partial^{2}y}{\partial t^{2}}\$
(for a longitudinal displacement s the same form holds).
This equation mathematically expresses that the shape of the disturbance propagates unchanged at speed v; the speed relation v = fλ follows directly by substituting a sinusoidal solution into the wave equation.
\$T\frac{\partial^{2}y}{\partial x^{2}} = \mu \frac{\partial^{2}y}{\partial t^{2}}\$
where μ is the linear mass density. Rearranging gives the speed
\$\boxed{v = \sqrt{\dfrac{T}{\mu}}}\$
\$B\frac{\partial^{2}s}{\partial x^{2}} = \rho \frac{\partial^{2}s}{\partial t^{2}}\$
where ρ is the density of the medium. Hence
\$\boxed{v = \sqrt{\dfrac{B}{\rho}}}\$
Goal: Demonstrate the difference between transverse and longitudinal waves.
Apparatus: (a) Long, taut string fixed at both ends, (b) Long, open-ended tube with a speaker, (c) Oscilloscope or microphone, (d) Signal generator.
Method (outline):
Analysis: Verify that the measured speed matches v = √(T/μ) for the string and v = √(B/ρ) for the air column, confirming the role of the different restoring forces.
| Feature | Transverse Wave | Longitudinal Wave |
|---|---|---|
| Particle motion | Perpendicular to the direction of propagation (up‑and‑down or side‑to‑side) | Parallel to the direction of propagation (back‑and‑forth) |
| Typical examples | Light in vacuum, waves on a string, surface water waves | Sound in air, pressure waves in a spring‑mass system, seismic P‑waves |
| Restoring force | Tension | Compression/expansion characterised by bulk modulus |
| Speed formula (for the simplest case) | \(v = \sqrt{T/\mu}\) (string) | \(v = \sqrt{B/\rho}\) (fluid/gas) |
| Polarisation | Can be polarised – the oscillation direction is defined. | Cannot be polarised – oscillation is along the travel direction. |
| Intensity (energy transport) | \(I = \tfrac12 \rho v \,\omega^{2} A^{2}\) (∝ A²) | \(I = \tfrac12 \rho v \,\omega^{2} A^{2}\) (same form, ρ is the medium density) |
| Typical diagram | Sinusoidal curve drawn perpendicular to the travel direction (crests & troughs). | Series of compressions and rarefactions drawn along the travel direction. |
For a wave travelling in the +x direction:
\$y(x,t)=A\sin\bigl(kx-\omega t+\phi\bigr)\$
The same functional form applies to longitudinal displacement s(x,t), with the particle motion now parallel to x.
\$I = I_{0}\cos^{2}\theta\$
where I₀ is the intensity before the analyser.
Source moving, observer stationary
\$f' = f\,\frac{v}{v\pm u_{\text{s}}}\$
Example: Police siren (f = 800 Hz) approaches a stationary observer at 30 m s⁻¹. With v = 340 m s⁻¹,
\$f' = 800\frac{340}{340-30}=800\frac{340}{310}\approx 880\ \text{Hz}\$
The observer hears a higher pitch while the source approaches and a lower pitch once it recedes.
| Region | Wavelength (λ) | Frequency (f) | Typical uses / examples |
|---|---|---|---|
| Radio | ≥ 1 m | ≤ 3 × 10⁸ Hz | Broadcasting, radar |
| Microwave | 1 mm – 1 m | 3 × 10⁸ – 3 × 10¹¹ Hz | Cooking, satellite communication |
| Infrared (IR) | 700 nm – 1 mm | 3 × 10¹¹ – 4 × 10¹⁴ Hz | Thermal imaging, remote sensing |
| Visible | 400 nm – 700 nm | 4 × 10¹⁴ – 7.5 × 10¹⁴ Hz | Human vision |
| Ultraviolet (UV) | 10 nm – 400 nm | 7.5 × 10¹⁴ – 3 × 10¹⁶ Hz | Sterilisation, fluorescence |
| X‑ray | 0.01 nm – 10 nm | 3 × 10¹⁶ – 3 × 10¹⁹ Hz | Medical imaging, crystallography |
| γ‑ray | < 0.01 nm | > 3 × 10¹⁹ Hz | Radioactive decay, astrophysics |
All EM waves are transverse, travel at speed c ≈ 3.00 × 10⁸ m s⁻¹ in vacuum, and can be polarised.
Q1 – Sound (longitudinal)
A sound wave of frequency 500 Hz travels through air where the speed of sound is 340 m s⁻¹.
Solution
\[
\lambda = \frac{v}{f} = \frac{340}{500}=0.68\ \text{m}
\]
Air particles oscillate back‑and‑forth along the direction of propagation, producing alternating compressions and rarefactions.
Q2 – String (transverse)
A string under a tension of 80 N has a linear mass density of 0.02 kg m⁻¹.
Solution
\[
v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80}{0.02}} = \sqrt{4000}=63.2\ \text{m s}^{-1}
\]
\[
\lambda = \frac{v}{f} = \frac{63.2}{250}=0.253\ \text{m}
\]
Particle motion is perpendicular to the string’s length (up‑and‑down).
Q3 – Polarisation (EM wave)
Unpolarised light of intensity 200 W m⁻² passes through two ideal polarisers whose transmission axes differ by 45°. Find the intensity after the second polariser.
Solution
First polariser reduces intensity to half: \(I_{1}=100\ \text{W m}^{-2}\).
Second polariser: \(I{2}=I{1}\cos^{2}45^{\circ}=100\times(0.707)^{2}=50\ \text{W m}^{-2}\).
Q4 – Doppler (sound)
A source emitting 600 Hz moves towards a stationary observer at 20 m s⁻¹. The speed of sound in air is 340 m s⁻¹. Calculate the frequency heard by the observer.
Solution
\[
f' = 600\frac{340}{340-20}=600\frac{340}{320}=637.5\ \text{Hz}
\]
(The observer hears a higher pitch because the source approaches.)
Transverse and longitudinal progressive waves share the fundamental relationships \(v = f\lambda\), the one‑dimensional wave equation, and an intensity that varies as the square of the amplitude. They differ in:
Understanding these differences, together with the Doppler effect for sound and the polarisation properties of electromagnetic waves, equips students to answer the full range of wave‑related questions in the Cambridge 9702 syllabus.
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