| 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. |
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:
\[
kx-\omega t+\phi = \text{constant}.
\]
\[
k\frac{dx}{dt}-\omega = 0\;\;\Longrightarrow\;\;\frac{dx}{dt}= \frac{\omega}{k}.
\]
\[
v = \frac{\omega}{k}.
\]
\[
v = f\lambda .
\]
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.
\[
K = \frac{1}{2}\mu\left(\frac{\partial y}{\partial t}\right)^{2}\Delta x .
\]
\[
U = \frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^{2}\Delta x .
\]
The instantaneous energy per unit length is therefore
\[
\mathcal{E}(x,t)=\frac{1}{2}\mu\left(\frac{\partial y}{\partial t}\right)^{2}
+\frac{1}{2}T\left(\frac{\partial y}{\partial x}\right)^{2}.
\]
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:
\[
I = \frac{\langle P\rangle}{A_{\text{cross}}}.
\]
| 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). |
The CRO provides a direct visual record of the time‑variation of the displacement (or a voltage proportional to it).
\[
T = n_{\text{div}}\;\Delta t,\qquad f = \frac{1}{T}.
\]
\[
A = \frac{V_{\text{pp}}}{2\;(\text{V/div})}\times\frac{1}{S},
\]
where \(S\) is the probe’s sensitivity (metres per volt).
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:
\[
v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{50}{0.020}} = 50\;\text{m s}^{-1}.
\]
\[
\omega = 2\pi f = 2\pi(25) = 157\;\text{rad s}^{-1}.
\]
\[
\langle P\rangle = \frac{1}{2}\mu v\omega^{2}A^{2}
= \frac{1}{2}(0.020)(50)(157)^{2}(2.0\times10^{-3})^{2}
\approx 0.25\;\text{W}.
\]
The calculation illustrates that even a modest amplitude can transport a measurable amount of energy when the frequency (hence \(\omega\)) is relatively high.
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