Download Solutions to Problems 23-25: String Equation, Quantum Mechanics, and Motion on a Cone and more Assignments Physics Fundamentals in PDF only on Docsity! Homework 6 Solutions Problems 23. A string of length 1.5 is suspended between the points (−0.5, 1.0) and (0.5, 1.0). (a) The equation for the string has the form y(x) = y0 + b cosh(x/b) and y′ = sinh(x/b). We want to determine b by requiring the length of the curve to be 1.5. The length of the curve is L = ∫ .5 −.5 dx √ 1 + (y′)2 = ∫ .5 −.5 dx √ 1 + (sinh(x/b))2 = ∫ .5 −.5 dx cosh(x/b) = 2b sinh(0.5/b) = 1.5 This is a transcendental equation for b, which can be solved graphically, or by using a program such as Mathematica or Excel. I used Maple to find b = 0.308. (b) Then to find y0, just plug back in for the end point x = 0.5, y = 1.0. The equation for the string gives 1.0 = y0 + b cosh(0.5/b). Solving for y0 with b = 0.308 gives y0 = 0.189. 24. The quantum mechanical wave function ψ(x) in an infinite-well potential satis- fies ψ(0) = ψ(a) = 0 and we can assume ψ is a real function for simplicity. We must minimize the expectation value of the kinetic energy E[ψ] = h̄ 2 2m ∫ a 0 dx (ψ′)2, while keeping the wave function normalized, ∫ a 0 dxψ2 = 1. (a) We have a global constraint 1− ∫ a 0 dxψ2 = 0. The functional to be mini- mized is I[ψ] = E[ψ]+λ [1− ∫ a 0 dxψ2] = ∫ a 0 dx [ h̄2 2m (ψ′)2 − λψ2 ] +λ, where λ is a Lagrange multiplier. (b) We can now apply the Euler-Lagrange equations to f = h̄ 2 2m (ψ′)2 − λψ2. We need ∂f ∂ψ′ = 2 h̄ 2 2m ψ′ and ∂f ∂ψ = −2λψ. Plugging into the Euler-Lagrange equations, ∂f ∂ψ − d dx ∂f ∂ψ′ = 0 gives −2λψ − 2 h̄ 2 2m ψ′′ = 0 or − h̄ 2 2m ψ′′ = λψ. This is just the time-independent Schroedinger Equation with no potential, and λ is the energy eigenvalue. (c) The most general form of the solution to the equation in (b) is ψ(x) = A cos(kx) +B sin(kx), where k = √ 2mλ/h̄. (At this point, the constant λ still could be negative or positive. The solution accommodates both possibilities.) (d) To make the wave function vanish at the origin ψ(0) = 0, requires A = 0. In order to vanish at the other side of the interval ψ(a) = 0 requires k = nπ/a, where n is an integer. Thus, there are a countably-infinite num- ber of local-minimum solutions (the eigenstates of the infinite-well). The Lagrange multiplier for each solution is the corresponding energy eigen- value, λ = h̄ 2n2π2 2ma2 . The constraint equation determines the final constant B by 1 = ∫ a 0 dxB2 sin2(kx) = ∫ a 0 dx 1 2 (1 − cos(2kx)) = B2a/2. Thus B = √ 2/a. 25. A particle of mass m is constrained to move on the surface of a cone of half angle α under the influence of gravity. The apex of the cone is at the origin, opening upward, and the axis of the cone is the z axis. (a) In cylindrical coordinates, the radius of the cone ρ is related to the height of the cone z by ρ/z = sinα/ cosα = tanα. We can then write the equation of constraint as ρ− z tanα = 0. (b) The Lagrange function for motion constrained to the cone is L = T −V +λ(ρ−z tanα), where the Lagrange multiplier λ(t) depends on time, since the constraint applies at all times. In cylindrical coordinates, we can write a general position as r = (ρ cos θ, ρ sin θ, z), and its time derivative as ṙ = ρ̇ûρ + ρθ̇ûθ + żûz, where ûρ, ûθ, and ûz are orthonormal basis vectors in cylindrical coordinates. The kinetic energy is T = 1 2 mṙ2 = 1 2 m(ρ̇2 + ρ2θ̇2 + ż2). The potential energy is just that due to gravity, V = mgz. In summary, L = 1 2 m(ρ̇2 + ρ2θ̇2 + ż2)−mgz + λ(ρ− z tanα). (c) Now find the equations of motion. Since ∂L ∂θ = 0, then ∂L ∂θ̇ = mρ2θ̇ is a conserved quantity. In fact, it is the angular momentum around the z- axis. Thus, the first equation is (1) mρ2θ̇ = constant = pθ. The other two equations are just straightforward applications of the Euler- Lagrange equations, so I will just write them down: (2) mz̈ = −mg − λ tanα. (3) mρ̈ = mρθ̇2 + λ. For later use, I will also write the equation of constraint again: (4) ρ = z tanα. (d) We now want an equation containing only the height z and its time- derivatives. We can use Eq. (3) to remove λ from Eq. (2), obtaining mz̈ +mg + (mρ̈−mρθ̇2) tanα = 0. We can then use Eq. (1) to remove the θ̇ dependence, and we can use the constraint Eq. (4) to remove the ρ dependence. We get mz̈(1 + tan2 α) +mg − p 2 θ mz3 tan2 α = 0. We can integrate this equation if we multiply both sides by ż, and note that z̈ż = d dt (ż2/2), ż = d dt z, and z−3ż = d dt (−z−2/2). Thus, the equation is a total derivative. Integrating gives 1 2 mż2(1 + tan2 α) +mgz + p2 θ 2mz2 tan2 α = constant = E. Note that this equation is actually just a conservation of energy of equa- tion, combined with the constraint equation and conservation of angular
