Assignment 5, solutions
Problem 1:
Consider a damped harmonic oscillator. Let us define T_{1} as the time between adjacent zero crossings, 2T_{1} as its “period”, and ω_{1} = 2π/(2T_{1}) 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  (8π^{2}n^{2})^{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 d^{2}x/dt^{2} + 2βdx/dt + ω_{0}^{2}x = 0. The solution for underdamped motion is x(t) = A exp(βt) cos(ω_{1}t  δ), with ω_{1}^{2} = ω_{0}^{2} – β^{2}. Here ω_{0} is the angular frequency of an undamped oscillator with the same spring constant. After n periods, nT = 2nT_{1}, the amplitude of the oscillator decreases to A/e. So exp(1) = exp(βnT), βnT_{ }= 1, β = 1/nT. With T = 2π/ω_{1} we have β = ω_{1}/(n2π). ω_{1}^{2} = ω_{0}^{2} – (ω_{1}/n2π.)^{2}, ω_{1}^{2} + ω_{1}^{2}/(4n^{2}π^{2}) = ω_{0}^{2}. ω_{1} = ω_{0} [1 + 1/(4n^{2}π^{2})]^{–½}. (1 + x)^{½} = 1 – ½ x + … . ω_{1} ≈ ω_{0} [1  1/(8n^{2}π^{2})]. 
Problem 2:
Consider the motion of a point of mass m subjected to a
potential energy function of the form
U(x) = U_{0}[1  cos(x/R_{0})]
for πR_{0}/2 < x < πR_{0}/2,
where x denotes distance, and U_{0} and R_{0} 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) = U_{0}/[1 + (x/R_{0})^{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 = (U_{0}/R_{0})sin(x/R_{0}), d^{2}U/dx^{2} = (U_{0}/R_{0}^{2})cos(x/R_{0}). dU/dx_{0} = 0. d^{2}U/dx^{2}_{0} > 0. (b) For x << R_{0} we have U(x) = ½(d^{2}U/dx^{2})_{0}x^{2}. This is the potential energy of a harmonic oscillator U(x) = ½kx^{2}, with k = d^{2}U/dx^{2}_{0}. (c) ω^{2} = (1/m)(d^{2}U/dx^{2})_{0} = U_{0}/(mR_{0}^{2}). T = 2π/ω = 2πm^{1/2}R_{0}/U_{0}^{1/2}. (d) dU/dx = (U_{0}/R_{0}^{2})2x/(1 + x^{2}/R_{0}^{2})^{2}, dU/dx_{0} = 0, x = 0 is an equilibrium position. d^{2}U/dx^{2}_{0} = (2U_{0}/R_{0}^{2}) > 0. For x << R_{0}, U(x) = U_{0}(1 – x^{2}/R_{0}^{2}) = U_{0} + (U_{0}/R_{0}^{2})x^{2.} ω^{2} = (2U_{0}/(mR_{0}^{2})). T = 2π/ω = 2^{1/2}πm^{1/2}R_{0}/U_{0}^{1/2}. 
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} + ℓ^{2}(dθ/dt)^{2} + 2(dX/dt)(dθ/dt)ℓcosθ]. U = mgℓcosθ + ½kX^{2}. L = T  U = ½(M + m)(dX/dt)^{2} + ½m[ℓ^{2}(dθ/dt)^{2} + 2(dX/dt)(dθ/dt)ℓcosθ] + mgℓcosθ  ½kX^{2}. 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)d^{2}X/dt^{2} + md^{2}θ/dt^{2}ℓcosθ  m(dθ/dt)^{2}ℓsinθ = kX. 2nd equation: ℓ^{2}d^{2}θ/dt^{2 }+ d^{2}X/dt^{2}ℓcosθ = 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.

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 q_{i}
(i = 0 to N  1) denote the displacement of the ith mass from its equilibrium
position. Assume periodic boundary conditions, q_{i} = q_{i+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 q_{j}(t) = Aexp(iφ_{j})
exp(iωt) = Aexp(i(φ_{j}  ωt)) exist, where ω is a normal mode
frequency.
Assume the phase of the amplitude depends on the position of the mass and write
φ_{j }= p*ja.
What are the restrictions on p due to the boundary conditions?
(d) Find the N normal mode frequencies ω_{n}. Make a sketch
of ω_{n} as a function of mode number n.
Solution:
Concepts: Lagrangian Mechanics, coupled oscillations, normal modes  
Reasoning: This is a onedimensional problem. We are asked to find the normal modes of coupled harmonic oscillators.  
Details of the calculation: (a) L = T  U = ∑_{i = 0}^{N1}[½m(dq_{i}/dt)^{2}  ½k(q_{i+1}  q_{i})^{2}] = ∑_{i = 0}^{N1}[½m(dq_{i}/dt)^{2}  ½k(q_{i+1}^{2} + q_{i}^{2}  2q_{i}q_{i+1}]. (b) d/dt(∂L/∂(dq_{j}/dt))  ∂L/∂q_{j} = 0. The terms in the Lagrangian that depend on q_{j }an dq_{j}/dt are ½m(dq_{j}/dt)^{2}  ½k(2q_{j}^{2}  2q_{j1}q_{j}  2q_{j}q_{j+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 md^{2}q_{j}/dt^{2}  k(q_{j+1}  2q_{j} + q_{j1}) = 0. (c) Assume q_{j}(t) = Aexp(i(p*ja  ωt). The periodic boundary conditions imply that p*ja = p*(j + N)a ± n*2π, p_{n}Na = n*2π, p_{n} = n2π/(Na), n = integer. (d) Inserting solutions of the form q_{j} = Aexp(i(p_{n}*ja  ωt)) into the equation of motion we obtain mω_{n}^{2} exp(ip_{n}*ja)  k[exp(ip_{n}*(j+1)a)  2exp(ip_{n}*ja) + exp(ip_{n}*(j1)a)] = 0, or ω_{n}^{2} = (k/m)[exp(ip_{n}*a)  2 + exp(ip_{n}*a)] = ¼ω_{0}^{2}[2cos(p_{n}*a)  2] = ½ω_{0}^{2}[1  cos(p_{n}*a)] = ω_{0}^{2}sin^{2}(p_{n}*a/2) = ω_{0}^{2}sin^{2}(nπ/N). Here ω_{0}^{2} = 2k/m. ω_{n} = ω_{0}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 twofold 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 L_{1} = 0.25 m, and the next resonance is heard when the length of the air column is L_{2} = 0.8 m. The speed of sound in air (at 20 ^{o}C, 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 oddinteger 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. L_{2}  L_{1} = 0.8 m  0.25 m = 0.55 m = λ/2, λ = 1.1 m. (b) f = v_{s(air)}/λ_{air} = (343 m/s)/(1.1 m) = 312 Hz. (c) λ_{water} = v_{s(water)}/f = (1490 m/s)/(312 Hz) = 4.8 m. 