Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / Section D

Fixed endpoints require $k_n=n\pi /L$, with $n=1,2,\dots$. Writing $c_s=\sqrt{T/\rho }$,For equal amplitudes $A$, the two requested profiles areThe first is asymmetric, with an interior zero at $x=2L/3$; the second has its interior node at $L/2$. For two-mode string recurrence, orthogonality means the whole initial profile recurs only when both cosine coefficients equal one. Since ${\omega }_2=2{\omega }_1$,

The weak-stiffness string frequency shift follows by expanding the supplied dispersion relation:Higher normal modes acquire larger fractional frequency shifts. Let $\eta =\alpha k_1^2/c_s^2\ll 1$ and retain ${\omega }_1^{(0)}=c_s{k}_1$ for the original ideal frequency. At the originally defined time $t_1=\pi /(2{\omega }_1^{(0)})$,Thus the $t=0$ sketch is unchanged, whereas the original quarter-period sketch gains a small negative fundamental component. If $t_1$ is redefined using the new fundamental frequency, its fundamental component is exactly zero and the second-mode change there is only second order. Both conventions agree that generic later shapes change because of relative phase drift.

The new frequency ratio isAn exact recurrence now requires ${\omega }_{1}t=2\pi q$ and ${\omega }_{2}t=2\pi p$ for integers $p,q$. If $r=p/q$ in lowest terms, the first recurrence is $2\pi q/{\omega }_1$; for irrational $r$ there is no exact finite recurrence. Near-recurrences remain possible. Positive stiffness alone does not prove that the ratio is irrational.

Here $V(x)$ is potential energy, $E$ is the stationary-state total energy eigenvalue, and $\psi (x)$ is its spatial wavefunction. Wavefunction normalization gives $\int |\psi {|}^{2}dx=1$, and $|\psi {|}^{2}dx$ is the probability of finding the particle in $dx$.

A bound state requires $-V_0<E<0$. Inside the well the equation is ${\psi }^{\prime \prime }=-k^2\psi$, whereas outside it is ${\psi }^{\prime \prime }=q^2\psi$, withThe ground state is even and has no node, selecting $\cos (kx)$ inside. Normalizability selects decaying, not growing, exterior exponentials. Continuity of $\psi$ and ${\psi }^{\prime }$ givesThe first excited state, if present, is odd: a sine inside and opposite-sign decaying tails outside. Matching gives $-k_1\cot (k_{1}a/2)=q_1>0$. On its first odd branch, $\pi /2<k_{1}a/2<\pi$, hence $k_{1}a>\pi$. Since $k_1^2<2m{V}_0/{\hslash }^2$, existence requiresThis is the first odd bound-state threshold of a square well. There is no excited bound state below that threshold; equality gives a non-normalizable zero-energy limiting state, not an extra bound state.

For the numerical well,The exterior probability of a symmetric finite-well state includes both tails:The threshold is $9.40\mathrm{eV}$, exceeding the well depth, so there is no excited bound state. The rounded supplied coefficients integrate to $0.9973$ rather than exactly one; this rounding does not change the quoted probability precision materially.