These are problems from the January 2024 qualifying exam. The address topics discussed in Physics 513. Problems 1 - 2 are from part 1 of the exam and problem 3 -4 are from part 2 of the exam.
Two objects are dropped from the top of a cliff of height H. The second object is dropped when the first has traveled a distance D. At the instant when the first object reaches the bottom, what is the distance of the second object above the ground in terms of H and D?
Solution:
In reference frame S a firecracker explodes and a second firecracker explodes a distance Δx = c*25 ns away and 52 ns later. In another inertial reference frame S', moving with velocity v i with respect to S, the two explosions are measured to occur a distance d' = c*42 ns apart in space. How much time passes between the explosions in frame S'?
Solution:
Consider a potential barrier with height U and width a.
(a) Which will tunnel more easily through the potential barrier, an electron or
a proton? Why? Assume each particle's kinetic energy is smaller than U.
(b) If the incoming particle's kinetic energy is 32.0 eV, U = 41 eV, and a =
0.25 nm, find the probability that the particle will tunnel through the barrier,
both for the proton and the electron.
Solution:
Algrebra:
C = ½exp((ik - ρ)a) (1 + ik/ρ) = ½exp(ika) exp(-ρa) + (ik/ρ)
½exp(ika) exp(-ρa).
C' = ½exp((ik + ρ)a) (1 - ik/ρ) = ½exp(ika) exp(ρa) - (ik/ρ)
½exp(ika) exp(ρa).
C + C' = ½exp(ika)[(exp(-ρa) + exp(ρa) + (ik/ρ)(exp(-ρa)
- exp(ρa))]
= exp(ika)[cosh(ρa) - (ik/ρ)(sinh(ρa)].
Similarly: C - C' = exp(ika)[-sinh(ρa) + (ik/ρ)(cosh(ρa)].
2A1 = A3 exp(ika)[ cosh(ρa) - (ik/ρ)sinh(ρa) + (ρ/ik)(-sinh(ρa) + (ik/ρ)(cosh(ρa)]
= A3 exp(ika)[2cosh(ρa) - (ik/ρ +ρ/ik) sinh(ρa)]
= A3 exp(ika)[2cosh(ρa) - i((k2-ρ2)/(ρk))sinh(ρa)].
A1 = ([(k2 - ρ2)/(2ikρ)]sinhρa +
coshρa)exp(ika)A3.
T = |A3/A1|2 = 4k2ρ2/[(k2 -
ρ2)2sinh2ρa + 4k2ρ2cosh2ρa]
= 4k2ρ2/[(k2 + ρ2)2sinh2ρa
+ 4k2ρ2].
[cosh2x - sinh2x = 1.]
T = 4E(U - E)/[U2sinh2[(2m(U - E)/ħ2)½a]
+ 4E(U - E)], or
T = (4E/U)(1 - E/U) / [sinh2[(2m(U0 - E)/ħ2)½ a]
+ (4E/U)(1 - E/U)].
If U, E, and a are the same, then the particle with the smaller mass (the
electron) will more easily tunnel through the potential barrier.
(b) Inserting numbers:
(4E/U)(1 - E/U) = 0.685.
For proton: (2m(U - E)/ħ2)½a = 165.
For the electron: (2m(U - E)/ħ2)½a = 3.84.
sinh(x) = ½(exp(x) - exp(-x)) ~ ½exp(x if x >> 1).
We can therefore write T ≈ (16E/U)(1 - E/U)exp(-2((2m/ħ2)(U - E))½a).
Tproton = 4*0.685*exp(-2*165) ≈ 1.3*10-143 ≈ 0.
Telectron = 4*0.685*exp(-2*165) ≈ 1.3*10-3 = 0.13%.
(a) For the system shown, consider only motion in the vertical plane.
Write the Lagrangian for small displacements from equilibrium of the system in
the form
where T and K are matrices. Clearly identify (write out) T and K.
If l1 = l2 = l, and m1 = 21 and m2 =
4 (everything in SI units), find the frequencies of the normal modes of this system? Briefly describe these modes.
Solution:
The potential energy of the system is
U = -m1gy1 - m2gy2 = (m1
+ m2)gl1θ2/2 + m2gl2φ2/2
+ constant.
The matrix U therefore is
(b) L = ½∑ij[Tij(dqi/dt)(dqj/dt)
- kijqiqj] with Tij = Tji,
kij = kji.
Solutions of the form qj = Re(Ajeiωt) can be
found. We can find the ω2 from det(kij - ω2Tij)
= 0.
With l1 = l2 = l and m1 = 21, m2 = 4 we have
ω2 = (25/21)(g/l) ± (10/21)(g/l).
ω12 = (35/21)(g/l) = (5/3)(g/l): φ = -(5/2)θ.
Ω22 = (15/21)(g/l) = (5/7)(g/l): φ = (5/2)θ.