Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 1 / Section D

For a small-slope string element, the transverse net tension is $T[y_x(x+\delta x)-y_x(x)]\simeq T{y}_{xx}\delta x$. Its mass is $\rho \delta x$, so Newton's second law givesIts extra length is $[\sqrt{1+y_x^2}-1]\delta x\simeq y_x^2\delta x/2$. Work against the approximately constant tension gives the elastic energy of a small-slope string, while transverse motion gives its kinetic energy:For a single travelling wave $y=f(x\pm ct)$, $y_t^2=c^2{y}_x^2$, so $dK=dU$ pointwise to this order. A standing wave need not have pointwise or instantaneous kinetic–potential equality.

For a fixed-end string of length L, taking the stated note to be its fundamental gives $c=2Lf$. HenceIf98Hz were instead the nth harmonic, the required tension would be $16.5/n^2\mathrm{N}$; the usual fundamental interpretation is explicit.

For the specified standing wave, $\dot{y}=A\omega \sin kx\cos \omega t$. Every allowed nonzero fixed-end mode has ${\int }_0^L{\sin }^{2}kxdx=L/2$, yieldingAt this instant the ideal mode's elastic energy is zero; at maximum displacement all its energy is elastic. ^[ref]

In the time-independent Schrödinger equation, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)$ is the kinetic-energy operator acting on the state, $V\psi$ its potential-energy contribution and $E\psi$ the energy-eigenvalue term. The complex wave function is a probability amplitude: the Born rule assigns density $|\psi {|}^2$ for a normalized single-particle state. A plane-wave beam uses an incident-amplitude or flux normalization instead of a unit-integral wavefunction.

For a free plane wave, ${\psi }^{\prime \prime }=-k^2\psi$, so $\boxed{E={\hslash }^2{k}^2/(2m)}$ and its time factor obeys $\hslash \omega =E$. Above the upward potential step, the bounded subthreshold solution has decay constantFor the PDF's unit incident amplitude, write $\psi =e^{ikx}+R{e}^{-ikx}$ on the left and $\psi =T{e}^{-\alpha x}$ on the right, omitting their common time phase. Continuity of both $\psi$ and ${\psi }^{\prime }$ at this finite step givesConsequentlyThe evanescent amplitude versus transmitted current distinction matters: the second quantity is a squared amplitude, not a transmitted-flux probability. The decaying tail has zero probability current, despite nonzero density and the possible value $|T{|}^2>1$.

Using $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and the electron mass and elementary charge specified above, the energy difference1eV gives $\alpha =5.12\mathrm{n}{\mathrm{m}}^{-1}$. Thus at the stated positionThis is relative to the unit incident density in the source figure, not an absolute normalized probability per metre. ^[ref]