Derive, using the definitions of speed, frequency and wavelength, the wave equation
\[
v = f\lambda
\]
and relate the other fundamental wave parameters (displacement, amplitude, phase‑difference, period, intensity) to the behaviour of a progressive wave.
| Parameter | Symbol | Definition | Typical Units |
|---|---|---|---|
| Displacement | \(y\) | Instantaneous distance of a particle from its equilibrium position. | m |
| Amplitude | A | Maximum displacement of a particle from equilibrium (peak of the wave). | m |
| Phase‑difference | \(\Delta\phi\) | Angular separation between two points of a wave; \(\displaystyle\Delta\phi = 2\pi\frac{\Delta x}{\lambda}=2\pi f\Delta t\). | rad |
| Period | T | Time for one complete cycle to pass a fixed point. | s |
| Frequency | f | Number of cycles per second; \(\displaystyle f = \frac{1}{T}\). | Hz |
| Wavelength | \(\lambda\) | Spatial distance between two consecutive points that are in phase (e.g. crest‑to‑crest). | m |
| Wave speed | v | Distance a given phase travels per unit time. | m s⁻¹ |
| Intensity | I | Power transmitted per unit area normal to the direction of propagation. | W m⁻² |

\[
v = \frac{\text{distance travelled}}{\text{time taken}} = \frac{x{2}-x{1}}{t{2}-t{1}} .
\]
\[
x{2}-x{1} = \lambda .
\]
\[
t{2}-t{1}=T .
\]
\[
v = \frac{\lambda}{T}.
\]
\[
v = \frac{\lambda}{1/f}=f\lambda .
\]
Thus, for any progressive wave,
\[
\boxed{v = f\lambda}
\]
\[
v = \sqrt{\frac{T}{\mu}} .
\]
\[
v = \sqrt{\frac{B}{\rho}} .
\]
\[
v = c = 3.00\times10^{8}\ \text{m s}^{-1}.
\]
\[
v = \frac{c}{\sqrt{\varepsilon_{r}}}.
\]
Mechanical (e.g. string, sound)
\[
P_{\text{avg}} = \tfrac12\,\mu\,\omega^{2}A^{2}v ,
\qquad \omega = 2\pi f .
\]
\[
I = \frac{P{\text{avg}}}{A{\perp}} \propto A^{2}.
\]
Electromagnetic (e.g. light, radio)
\[
I = \frac{1}{2}\,c\,\varepsilon{0}\,E{0}^{2}
= \frac{1}{2}\,\frac{c}{\mu{0}}\,B{0}^{2},
\]
where \(E{0}\) and \(B{0}\) are the peak electric and magnetic field amplitudes.
\[
T = (\text{number of divisions})\times(\text{time per division}).
\]
Problem: A sound wave travels in air at \(v = 340\ \text{m s}^{-1}\) and has a frequency of \(f = 500\ \text{Hz}\). Find its wavelength and intensity, given a pressure‑amplitude \(A = 2.0\times10^{-5}\ \text{Pa}\). (For sound in air, \(I = p_{\text{rms}}^{2}/(\rho c)\) with \(\rho = 1.2\ \text{kg m}^{-3}\) and \(c = v\).)
\[
\lambda = \frac{v}{f}= \frac{340}{500}=0.68\ \text{m}.
\]
\[
p_{\text{rms}} = \frac{A}{\sqrt{2}} = 1.41\times10^{-5}\ \text{Pa}.
\]
\[
I = \frac{p_{\text{rms}}^{2}}{\rho c}
= \frac{(1.41\times10^{-5})^{2}}{1.2 \times 340}
\approx 4.9\times10^{-13}\ \text{W m}^{-2}.
\]
Result: \(\lambda = 0.68\ \text{m}\), \(I \approx 5\times10^{-13}\ \text{W m}^{-2}\).
Answer: \(\lambda\) is halved because \(\lambda = v/f\).
Answer: The equation follows solely from the definitions of speed (distance / time), frequency (cycles / time) and wavelength (distance per cycle). Whether the particle motion is perpendicular (transverse) or parallel (longitudinal) to the direction of travel, one complete cycle still travels a distance \(\lambda\) in a time \(T = 1/f\), giving the same relationship.
Answer: Since \(I \propto A^{2}\), the intensity increases by a factor of \(3^{2}=9\).
The parameters introduced here—\(v, f, \lambda, A, I\) and phase—are reused throughout the wave chapter. In the next sections you will:
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