Past exam of the mathematical methods II course of the University of Oxford / 2021 resit (A10289W1)

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1

The examination date is 9 September 2021. The source archive labels its member 2020/L/a10289w1.pdf; the date on the paper establishes the year. Solutions cover all questions, including every option in Section B.

The Maclaurin series of the exponential function and sine giveBoth the numerator and denominator first become nonzero at order $x^4$. Dividing by that common power gives the limit

The fundamental theorem of calculus differentiates the variable upper endpoint:At zero, $I(0)=0$, $I^{\prime }(0)=2/\sqrt{\pi }$, $I^{\prime \prime }(0)=0$ and $I^{\prime \prime \prime }(0)=-4/\sqrt{\pi }$. The integrand is even, so its integral from zero is an odd function; all even coefficients in the Maclaurin series vanish. HenceThis is the local expansion of the error function.

First method: substitute the trajectory. Along the parametric curve, the temperature is $T(t)=20e^{-{\cos }^{2}t}-(1+\sin t{)}^3$. The chain rule givesSecond method: use the spatial derivatives. The partial derivatives of the scalar field are $T_x=-40x{e}^{-x^2}$ and $T_z=-3(z+1{)}^2$, while the velocity components are $\dot{x}=-\sin t$ and $\dot{z}=\cos t$. The multivariable chain rule therefore giveswhich agrees with the first calculation. This is the gradient dotted with the velocity, rather than a partial derivative at a fixed position.

Apply the product-to-sum formula $2\cos u\cos v=\cos (u+v)+\cos (u-v)$ to the amplitude modulation:The amplitude modulation sidebands have angular frequencies $\omega +{\omega }_m$ and $\omega -{\omega }_m$, each with amplitude $b/2$ in this real cosine convention. The carrier wave has angular frequency $\omega$ and amplitude $a$.

This identity is exact for the supplied space-time expression. A physical qualification is needed if every term is meant to be a freely propagating electromagnetic wave with speed of light $c$: the dispersion relation requires $k_{\pm }=(\omega \pm {\omega }_m)/c$, whereas the prescribed modulation at every position leaves all three wavenumbers equal to $k=\omega /c$. A freely propagating modulated wave instead has envelope $a+b\cos ({\omega }_m(t-x/c))$; its analogous expansion has $k_{\pm }$. For example, at $x=t=0$ the prescribed expression has $A_{tt}-c^2{A}_{xx}=-b{\omega }_m^2$, which is nonzero for nontrivial modulation. The given formula can describe modulation imposed throughout space or the temporal signal at one position.

Differentiate the parametric curve:The arc length element is $ds=\sqrt{{\dot{x}}^2+{\dot{y}}^2}dt=|a|tdt$ on $0\le t\le 2$. ThusFor a positive geometric scale $a$, this is $2a$. The absolute value is necessary when the unspecified real constant $a$ is negative; $a=0$ gives zero arc length.

The gradient of the scalar field isFor a unit vector $\hat{\mathbf{n}}$, the directional derivative is $D_{\hat{\mathbf{n}}}f=\nabla f\cdot \hat{\mathbf{n}}$. The Cauchy-Schwarz inequality bounds its magnitude by $\Vert \nabla f\Vert =2\sqrt{3041}$, with equality precisely in the two directions parallel to the gradient. ThereforeThe plus choice gives the greatest positive directional derivative, and the minus choice gives the most negative directional derivative. Both maximize its magnitude.

For the scalar surface integral over an obliquely cut cylinder, split the scalar surface integral into the curved cylinder and its two planar caps. Write $s=z-x$, so the solid has $x^2+y^2\le 4$ and $-3\le s\le 2$. On the curved cylinder, use the parametrizationThe surface area element follows fromConsequently the curved contribution isOn either cap $z=x+s$, the surface area element is $\sqrt{1+z_x^2+z_y^2}dxdy=\sqrt{2}dxdy$ over a radius-two disk. The lower cap has $x-z=3$ and the upper cap has $x-z=-2$. Thusand the complete scalar surface integral isThe surface area element is unsigned: opposite outward normals on the caps do not introduce extra minus signs.

The probability density function is nonnegative and vanishes on negative lifetimes. Since $0.04=1/25$, its normalization of a probability distribution isIt is an exponential distribution with rate $1/25$ per month.

Integrating the probability density function over the stated interval givesThe strict endpoint does not affect the probability for this continuous probability distribution.

The survival function of this exponential distribution is $\mathbb{P}(T>t)=e^{-t/25}$ for $t\ge 0$. Therefore

Subtract the two values of the survival function, or integrate the probability density function directly:The final zero records the requested third significant figure.

The expectation follows by integration by parts, with $u=t$ and $dv=e^{-t/25}dt/25$:The endpoint term vanishes because the exponential function decays faster than $t$ grows. Hence the mean lifetime is

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Take $0<r_1<r_2$, so the shell excludes the possible singularity at the origin. The divergence and curl of a radial vector field formula in spherical coordinates givesWith volume element $dV=r^2\sin \theta drd\theta d\varphi$, direct integration givesFor the independent boundary calculation, the outer sphere has outward unit normal vector $+\hat{\mathbf{r}}$ and the inner sphere has outward unit normal vector $-\hat{\mathbf{r}}$. Their fluxes areTheir sum equals the direct volume integral, verifying the divergence theorem. The restriction $\beta \ne 2$ permits the displayed power-antiderivative step. At $\beta =2$, treated separately, both the shell volume integral and net boundary flux are zero; the excluded origin must still not be filled in.

Away from the origin, write $\mathbf{F}=r^{-\beta -1}\mathbf{r}$. The curl product rule for a scalar field and the gradient of a radial function giveConsequently the surface integral of curl over the upper hemisphere is zero. Its outward unit normal vector induces a counterclockwise equator when viewed from above. Parametrize that boundary byThe radial vector field is perpendicular to this tangent, soThis verifies Stokes theorem directly without passing through the singular origin.

Regard the planar vector field as $(A(x,y),B(x,y),0)$, with no dependence on $z$. Its curl isThus, for continuously differentiable $A$ and $B$, the necessary and sufficient pointwise condition isTo deduce global path independence from this local condition also requires a suitable domain, for example a simply connected domain.

Set $P=10x^4-2x{y}^3$ and $Q=-3x^2{y}^2$. Their partial derivatives satisfy $P_y=-6x{y}^2=Q_x$ throughout ${\mathbb{R}}^2$. The vector field is smooth on the entire simply connected plane, so zero curl proves path independence.

Choose the broken straight path $(0,0)\to (2,0)\to (2,1)$. Along its horizontal segment $y=0$ and $dy=0$, givingAlong its vertical segment $x=2$ and $dx=0$, givingTherefore the line integral is

An exact differential is an expression $Pdx+Qdy$ that equals $df=f_{x}dx+f_{y}dy$ for a single-valued differentiable function $f$ on the domain. Its line integral is an endpoint difference by the fundamental theorem for line integrals.

Integrating $f_x=P$ with respect to $x$ gives $f=2x^5-x^2{y}^3+g(y)$. Matching $f_y=-3x^2{y}^2+g^{\prime }(y)$ to $Q$ requires $g^{\prime }(y)=0$. HenceThe arbitrary additive constant cancels. This constructs the vector-field potential globally and provides a second proof of path independence.

Assume a simple, piecewise smooth closed curve oriented counterclockwise around its enclosed region $D$. Apply Green's theorem with $P=-y$ and $Q=x$:The orientation assumption is essential. A clockwise unit circle, $x=\cos \theta$, $y=-\sin \theta$ for $0\le \theta \le 2\pi$, gives $-2\pi$ although its enclosed area is $\pi$. For a self-intersecting curve the same line integral measures twice the signed area, with multiplicity, rather than necessarily twice an ordinary enclosed area.

For positive semiaxes $a,b$, the ellipse has the counterclockwise parametrization $x=a\cos \theta$, $y=b\sin \theta$. ThenThereforeIf nonzero signed constants are allowed in place of positive semiaxes, the geometric area is $\pi |ab|$.

Put $\xi =x+vt$ and $F(\xi )=A{e}^{-{\xi }^2/x_0^2}$, with $x_0\ne 0$. The chain rule givesFor a nonzero pulse the wave equation on a string therefore requiresIt is not a solution for arbitrary $v$; the zero pulse $A=0$ is the trivial exception. Here $\mu$ is linear mass density, and $T>0$ is tension. With $v>0$, the maximum lies at $x=-vt$, so the Gaussian function translates towards decreasing $x$ by $v(t_2-t_1)$ between the two times. Its amplitude and width stay constant.

A reference string segment $dx$ has mass $\mu dx$ and transverse velocity $y_t$. Its kinetic energy is $dK=\mu y_t^{2}dx/2$. In the small-slope approximation its extra arc length isThe work against constant tension $T$ supplies the corresponding elastic potential energy $dU=T(d\ell -dx)$. To quadratic order, the energy density per unit reference length is thereforeFor the travelling wave $y=A\cos (kx-\omega t+\varphi )$, the wave equation implies ${\omega }^2=(T/\mu )k^2$. Its kinetic energy density and elastic energy density are equal, soThe energy per wavelength of a transverse string wave comes from integration over any full wavelength $\lambda =2\pi /|k|$:The printed expression for energy per wavelength is missing a factor $\lambda$. Its right-hand side is the mean energy density, with units of energy per length. For a concrete counterexample within the small-slope regime, take $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $T=1\mathrm{N}$, $A=0.01\mathrm{m}$ and $\omega =\pi {\mathrm{s}}^{-1}$. Then $\lambda =2\mathrm{m}$ and direct integration gives $E_{\lambda }=10^{-4}{\pi }^2\mathrm{J}$, whereas the printed right-hand side is $5\times 10^{-5}{\pi }^2\mathrm{J}{\mathrm{m}}^{-1}$. It cannot equal an energy.

For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.

Use signed arc length displacements $x_1,x_2$ in the common clockwise sense shown by both arrows in the source diagram. Near the diametrically opposite equilibrium, one spring changes length by $x_2-x_1$ and the other by $x_1-x_2$. Their combined elastic potential energy isThe kinetic energy is $K=(m_1{\dot{x}}_1^2+q{m}_1{\dot{x}}_2^2)/2$. The tangential components of Newton's second law, or the derivatives $-\partial U/\partial x_i$, giveBoth springs contribute to each restoring force. The hoop radius drops out because the coordinates are arc lengths; angular coordinates $x_i/r$ give the same equations after cancelling the common factor $r^2$ in the energies.

To find the normal modes of two arc-spring-coupled beads on a hoop, take $x_i=a_i{e}^{i\Omega t}$ for a normal mode. The resulting generalized eigenvalue problem isIts determinant isThus the angular frequencies areThe ordinary nonzero frequency is $\omega /(2\pi )$. The zero mode has mode shape $(1,1)$ and represents a common rotation with no change in spring length. The oscillatory mode shape is $(1,-1/q)$.

Define a common displacement coordinate $X=(x_1+q{x}_2)/(1+q)$ and a relative displacement $\delta =x_1-x_2$. Adding and subtracting the equations of motion givesThe general real motions of the two normal modes are thereforeHere $C,D,A,B$ are independent real constants. The zero mode includes uniform drift as well as a static common displacement; writing only a constant times $\cos 0t$ would miss the independent linear-in-time solution. Their superposition isIn the oscillatory normal mode, $m_1{x}_1+m_2{x}_2=0$ and the two amplitudes have ratio $1/q$, with opposite phases. Equivalently the oscillatory motion has zero total angular momentum about the hoop center.

As $q\to \infty$ with $m_1,k$ fixed, $\omega \to \sqrt{2k/m_1}$ and the heavier particle's oscillatory amplitude tends to zero for fixed lighter-particle amplitude. The lighter particle then oscillates against two effectively anchored springs with combined spring constant $2k$. A common uniform rotation remains an allowed zero mode; the heavy particle is stationary in the limit only if the initial common drift is zero or tends to zero, as it does for bounded total angular momentum.

For $q=2$, the common coordinate has $X(0)=(x_1(0)+2x_2(0))/3=0$ and $\dot{X}(0)=\pi \omega r/20$. The relative coordinate has $\delta (0)=3\pi r/20$ and $\dot{\delta }(0)=0$. ConsequentlySince $x_1=X+2\delta /3$ and $x_2=X-\delta /3$, the full superposition isThe heavier particle's velocity isIt first vanishes when $\sin \omega t=-1$, soThis velocity is nonnegative and only touches zero: the heavier particle does not reverse its direction at that instant. Both the uniform zero mode and the oscillatory normal mode are needed to obtain this result.