Past exam of the physics A course of the University of Cambridge / 2013 / Paper 1 / Section A

In the small-angle approximation the physical pendulum is a harmonic oscillator with $\omega =2\pi /T$. Its classical energy is $E=m{\omega }^2{A}^2/2=3.55\times 10^{-3}\mathrm{J}$. Equating this to $(n+1/2)\hslash \omega$ givesAn oscillating bob is not a single stationary energy eigenstate. An ideal coherent state has a Poisson distribution of oscillator occupations centred on this mean, with relative spread $1/\sqrt{\bar{n}}$, exceptionally small here. This quantum occupation scale of a classical oscillator fixes the excitation scale without requiring the real, damped clock to be a pure coherent state.

The free particle dispersion relation is $\hslash \omega ={\hslash }^2{k}^2/(2m)$. HenceThe phase velocity follows constant-phase surfaces of a plane wave. The group velocity follows the envelope of a narrow-band wave packet and equals its classical velocity $p/m$. Dispersion can spread the packet; the phase speed is not the particle's transport velocity. In three dimensions ${\mathbf{v}}_g=\hslash \mathbf{k}/m$ and the phase speed quoted is along $\mathbf{k}$.

The mass dependence of fixed-barrier tunnelling follows by holding energy and the potential-energy barrier fixed: quantum tunnelling has the leading WKB approximationThe alpha-particle mass is about $7.3\times 10^3$ electron masses, so its decay exponent is about 85 times larger for the same $V-E$ profile. Exponential suppression therefore makes transmission much smaller. This comparison holds the potential energy and incident energy fixed: the same electrostatic potential is not the same potential-energy barrier for charges of different magnitude and sign.

For the squared discrepancy of a digit frequency, write $d_N^2=(f_0-0.1{)}^2$. The plotted ordinates are $10^{-2}$, $4.00\times 10^{-4}$, $1.69\times 10^{-4}$, $4.90\times 10^{-5}$, $3.60\times 10^{-5}$ and $9.00\times 10^{-6}$ in increasing $N$. Under an independent, equiprobable-digit model, the zero count is binomial andThus the expected log-log slope is $-1$. If “relative” means division by the expected fraction before squaring, every ordinate is multiplied by 100 and the prediction becomes $9/N$; the slope is unchanged. These are nested prefixes of one digit sequence, so the points are correlated, rounded observations, not independent variance estimates. Their rough trend supports a reciprocal-power scale but does not prove independence or that $\pi$ is a normal number.

From Snell's law, $n=\sin {\theta }_i/\sin {\theta }_r=1.5204$. Treating the quoted angle errors as independent one-standard-deviation uncertainties and using radians in derivatives,This refractive-index uncertainty from Snell's law is first-order propagation of uncertainty, with air's index approximated by one. Correlated angular calibration errors add $-2n^2\cot {\theta }_i\cot {\theta }_r\mathrm{Cov}({\theta }_i,{\theta }_r)$. If the quoted errors are hard bounds instead, their conservative first-order sum gives about $0.19$, rather than the statistical quadrature value.