Assignment 5, solutions

Problem 1:

Consider a damped harmonic oscillator.  Let us define T1 as the time between adjacent zero crossings, 2T1 as its “period”, and ω1 = 2π/(2T1) as its “angular frequency”.  If the amplitude of the damped oscillator decreases to 1/e of its initial value after n periods, show that the frequency of the oscillator must be approximately [1 - (8π2n2)-1] times the frequency of the corresponding undamped oscillator.

Solution:

 Concepts:The underdamped harmonic oscillator Reasoning:The oscillator is underdamped, since it crosses the equilibrium position many times. Details of the calculation:The equation of motion for the damped harmonic oscillator is d2x/dt2 + 2βdx/dt + ω02x = 0.The solution for underdamped motion is x(t) = A exp(-βt) cos(ω1t - δ), with ω12 = ω02 – β2.  Here ω0 is the angular frequency of an undamped oscillator with the same spring constant.After n periods, nT = 2nT1, the amplitude of the oscillator decreases to A/e. So exp(-1) = exp(-βnT),  βnT = 1,  β = 1/nT.With T = 2π/ω1 we have β = ω1/(n2π). ω12 = ω02 – (ω1/n2π.)2,   ω12 + ω12/(4n2π2) = ω02.ω1 = ω0 [1 + 1/(4n2π2)]–½.(1 + x)-½ = 1 – ½ x + … .ω1 ≈ ω0 [1 - 1/(8n2π2)].

Problem 2:

Consider the motion of a point of mass m subjected to a potential energy function of the form
U(x) = U0[1 - cos(x/R0)] for  πR0/2 < x < πR0/2,
where x denotes distance, and U0 and R0 are positive constants with dimensions of energy and length, respectively.
(a)  Find the position of stable equilibrium for the mass.
(b)  Show that the motion of the mass in proximity of the stable equilibrium position is SHM.
(c)  Find the period of the small oscillations.
(d)  Find the period of the small oscillations for the same mass in the potential
U(x) = -U0/[1 + (x/R0)2].

Solution:

 Concepts:Oscillations about an equilibrium position Reasoning:We are asked to find the period of small oscillations. Details of the calculation:(a)  The position of stable equilibrium is x = 0.dU/dx = (U0/R0)sin(x/R0),  d2U/dx2 = (U0/R02)cos(x/R0).dU/dx|0 = 0.  d2U/dx2|0 > 0.(b)  For x << R0 we have U(x) = ½(d2U/dx2)|0x2.  This is the potential energy of a harmonic oscillator U(x) = ½kx2, with k = d2U/dx2|0.  (c)  ω2 = (1/m)(d2U/dx2)|0 = U0/(mR02).  T = 2π/ω = 2πm1/2R0/U01/2. (d)  dU/dx = (U0/R02)2x/(1 + x2/R02)2,  dU/dx|0 = 0, x = 0 is an equilibrium position.d2U/dx2|0 = (2U0/R02) > 0.For x << R0, U(x) = -U0(1 – x2/R02) = -U0  + (U0/R02)x2.ω2 = (2U0/(mR02)).  T = 2π/ω = 21/2πm1/2R0/U01/2.

Problem 3:

A pendulum consisting of a mass m and a weightless string of length l is mounted on a mass M, which in turn slides on a support without friction and is attached to a horizontal spring with force constant k, as seen in the diagram.  There is a slot in the support in order that the pendulum may swing freely.
(a)  Set up Lagrange's equations.
(b)  Find the normal mode frequencies for small oscillations. What are those frequencies to zeroth order in m/M, when  m << M?

Solution:

 Concepts:Lagrangian Mechanics, coupled oscillations Reasoning:We are asked to write down and solve Lagrange's equations. Details of the calculation:(a) T = ½M(dX/dt)2 + ½m[(dx/dt)2 + (dy/dt)2],  x = X + ℓ sinθ, y = -ℓcosθ. T = ½M(dX/dt)2 + ½m[(dX/dt)2 + ℓ2(dθ/dt)2 + 2(dX/dt)(dθ/dt)ℓcosθ]. U = -mgℓcosθ + ½kX2.  L = T - U = ½(M + m)(dX/dt)2 + ½m[ℓ2(dθ/dt)2 + 2(dX/dt)(dθ/dt)ℓcosθ] + mgℓcosθ - ½kX2.  We have two generalized coordinates, X and θ.  Lagrange's equations are d/dt ∂L/∂(dX/dt) - ∂L/∂X = 0,  d/dt ∂L/∂(dθ/dt) - ∂L/∂θ = 0. 1st equation:  (M + m)d2X/dt2 + md2θ/dt2ℓcosθ - m(dθ/dt)2ℓsinθ = -kX. 2nd equation: ℓ2d2θ/dt2 + d2X/dt2ℓcosθ = -gℓ sinθ.(b)  Assume X and θ are small quantities, and in the equations of motion keep only terms to first order in small quantities.  Then we have, using sinθ ~ θ, cosθ ~ 1. 1st equation:  (M + m)d2X/dt2 + mℓd2θ/dt2  = -kX. 2nd equation: d2θ/dt2 + (1/ℓ)d2X/dt2  = -(g/ℓ)θ. We obtain the same equations if we make the small displacements approximation earlier and keep only terms up to second order in the Lagrangian. Try solutions X = Aexp(iωt),  θ = (B/ℓ)exp(iωt),  then -ω2(M + m)A - ω2mB = -kA,  -ω2B - ω2A = -(g/ℓ)B. We solve this system of equations by setting   ω4 – ω2(k/M + (g/ℓ)(1 + m/M)) + kg/(ℓM) = 0. ω2 = (k/M + (g/ℓ)(1 + m/M))/2 ± ½((k/M + (g/ℓ)(1 + m/M))2 - 4kg/(ℓM))½. ω+2 = (k/M + (g/ℓ)(1 + m/M))/2 + ½((k/M + (g/ℓ)(1 + m/M))2 - 4kg/(ℓM))½. ω-2 = (k/M + (g/ℓ)(1 + m/M))/2 - ½((k/M + (g/ℓ)(1 + m/M))2 - 4kg/(ℓM))½. The frequencies for small oscillations are ω+ and ω-. [For the relative displacements we have B =  Aω2/((g/l) - ω2).  ω+2 > g/l,  so A and B have opposite signs for this mode.] As m --> 0, ω+ --> (k/M)½, ω- = (g/ℓ)½.  The modes effectively decouple.  The small mass m does no longer influence the motion of the big mass M.  When M oscillates with frequency (k/M)½, m becomes a driven harmonic oscillator without damping (for small oscillations).  The most general solution is a superposition of oscillations with the driving frequency and oscillations with the natural frequency of the oscillator.  (See Problem 3 of the problems worked in class.)

Problem 4:

A large number N (N = even) of point masses m are connected by identical springs of equilibrium length a and spring constant k.  Let qi (i = 0 to N - 1) denote the displacement of the ith mass from its equilibrium position.  Assume periodic boundary conditions, qi = qi+N.  (You can, for example imagine the masses arranged on a large circle of circumference Na.)
(a)  Write down the Lagrangian for the system of N point masses.
(b)  Find the equation of motion for the jth point mass.
(c)  Assume solutions of the form qj(t)  = |A|exp(iφj) exp(-iωt)  = |A|exp(i(φj - ωt)) exist, where ω is a normal mode frequency.
Assume the phase of the amplitude depends on the position of the mass and write φj = p*ja.
What are the restrictions on p due to the boundary conditions?
(d)  Find the N normal mode frequencies ωn.  Make a sketch of ωn as a function of mode number n.

Solution:

 Concepts:Lagrangian Mechanics, coupled oscillations, normal modes Reasoning:This is a one-dimensional problem.  We are asked to find the normal modes of coupled harmonic oscillators. Details of the calculation: (a)  L = T - U = ∑i = 0N-1[½m(dqi/dt)2 - ½k(qi+1 - qi)2] = ∑i = 0N-1[½m(dqi/dt)2 - ½k(qi+12 + qi2 - 2qiqi+1]. (b)  d/dt(∂L/∂(dqj/dt)) -  ∂L/∂qj = 0. The terms in the Lagrangian that depend on qj an dqj/dt are ½m(dqj/dt)2 - ½k(2qj2 - 2qj-1qj - 2qjqj+1). (They come from the terms in the sum i = j and i + 1 = j.) The equation of motion for the jth point mass therefore is md2qj/dt2 - k(qj+1 - 2qj + qj-1) = 0. (c)  Assume qj(t)  = |A|exp(i(p*ja - ωt).  The periodic boundary conditions imply that p*ja = p*(j + N)a ± n*2π,  pnNa = n*2π,  pn = n2π/(Na),  n = integer. (d)  Inserting solutions of the form qj = |A|exp(i(pn*ja - ωt)) into the equation of motion we obtain -mωn2 exp(ipn*ja) - k[exp(ipn*(j+1)a) - 2exp(ipn*ja) + exp(ipn*(j-1)a)] = 0, or ωn2 = -(k/m)[exp(ipn*a) - 2 + exp(-ipn*a)] = -¼ω02[2cos(pn*a) - 2] = ½ω02[1 - cos(pn*a)] = ω02sin2(pn*a/2) = ω02sin2(nπ/N). Here ω02 = 2k/m. ωn = ω0|sin(nπ/N)| are the normal mode frequencies. There are N frequencies which we can label with a mode number from n = -N/2 to (N/2 -1). All but two of the frequencies are two-fold degenerate.

Problem 5:

A vibrating tuning fork is held above a column of air as shown in the diagram.  The reservoir is raised and lowered to change the water level, and therefore the length of the column of air.  The shortest length of air column that produces a resonance is L1 = 0.25 m, and the next resonance is heard when the length of the air column is L2 = 0.8 m.  The speed of sound in air (at 20 oC, the temperature during the experiment) is 343 m/s and the speed of sound in water is 1490 m/s.

(a)  Calculate the wavelength of the standing sound wave produced by the tuning fork.
(b)  Calculate the frequency of the tuning fork that produces the standing wave.
(c)  Calculate the wavelength of the sound wave produced by this tuning fork in water.

Solution:

 Concepts: Standing waves Reasoning: For the longest wavelength standing wave in a tube of length L with one open end and one closed end (neglecting edge effects) we have L = λ/4.  For the next longest standing wave we have L = 3λ/4. [An odd-integer number of quarter wavelengths have to fit into the tube of length L. L = nλ/4,  λ = 4L/n,  f = v/λ  = nv/(4L),  n = odd.] Details of the calculation: (a)  In an experiment with a tube we always have edge effects.  We therefore look at the difference in tube length between the third harmonic and the fundamental for the same frequency. L2 - L1 = 0.8 m - 0.25 m = 0.55 m = λ/2,  λ = 1.1 m. (b)  f = vs(air)/λair = (343 m/s)/(1.1 m) = 312 Hz. (c)  λwater = vs(water)/f = (1490 m/s)/(312 Hz) = 4.8 m.