Review Problems

Problem 1:

The S-shaped wire shown has a mass M, and the radius of curvature of each half is R.  It lies in the yz-plane.
Let the gravitational acceleration g point downward.
(a)  Find its moment of of inertia about an axis parallel to the x-axis passing through point A.
(b)  Find its moment of of inertia about an axis parallel to the x-axis passing through point B.
(c)  If the wire is allowed to swing in the yz-plane with the pivot at point B, find the frequency of small oscillations

Solution:

• Concepts:
  The moment of inertia, the parallel axis theorem
• Reasoning:
  The parallel-axis theorem states that if I_CM is the moment-of-inertia of an object about an axis through its center of mass, then I, the moment of inertia about any axis parallel to that first one is given by I = I_CM + md², where m is the object's mass and d is the perpendicular distance between the two axes.
• Details of the calculation:
  (a)  From the definition of the moment of inertia about an axis, I = ∑m_ir_i², where r_i is the perpendicular distance of particle i from the axis, we see that the moment of inertia of the wire about an axis through A is the same as that of a complete circular loop about an axis through A.  The moment of inertia of a loop through its center is I_center = MR².  Using the parallel axis theorem we therefore have for the wire
  I_A = MR² + MR² = 2MR².
  (b)   The moment of inertia of the wire about its CM is 2MR².  The CM is a parallel distance of 2R from the point B.
  Using the parallel axis theorem we therefore have for the wire
  I_B = 2MR² + 4MR² = 6MR².
  (c)  For small oscillations about point B we have
  T = ½I(dθ/dt)² = 3MR²(dθ/dt)².  U = -Mg2R(1 - θ²/2)
  6MR²d²θ/dt² = -2MgRθ.   d²θ/dt² = -(1/3)(g/R)θ.  ω² = g/(3R).  f = ω/(2π).

Problem 2:

In one dimension, a particle has a wave function ψ(x) = d^-½ exp(-|x|/d) at t = 0.
(a)  Is ψ(x) normalized?
(b)  What is the probability that a measurement of the position of the particle at t = 0 will yield a result between x₁ and x₂ (x₂ > x₁ > 0)?
(c)  Can this wave function be the eigenfunction of some Hamiltonian H(x), i.e. can it represent a stationary state?

Solution:

• Concepts:
  The fundamental assumptions of quantum mechanics
• Reasoning:
  The wave function ψ(r,t) is interpreted as a probability amplitude of the particles presence.  |ψ(r,t)|² is the probability density. 
• Details of the calculation:
  (a)  Probability density:  P(x)dx = |ψ(x)|²dx.
  (2/d)∫₀^∞exp(-2x/d)dx = (2/d)(d/2) = 1.
  ψ(x) is normalized?
  (b)  P (x₂ > x > x₁) = (1/d)∫_-x1^x2 exp(-2x/d) dx = ½∫_-2x1/d^2x2/d exp(-y) dy
  = ½(exp(-2x₁/d) - exp(-2x₂/d)).
  (c)  This wave function could represent a stationary state, if the Hamiltonian H = p²/(2m) + U(x), with U(x) proportional to -δ(x).

Problem 3:

In this one-dimensional problem, a cart of mass M can move horizontally along the x-axis.  A mass m is attached to the car by a spring with spring constant k.  Let X denote the equilibrium position of the end of the spring attached to m, and x the displacement of the spring from its equilibrium position.  Neglect friction.
(a)  Write down the Lagrangian of the system.
(b)  Find the equations of motion.
(c)  What are the normal mode frequencies of the system?

Solution:

• Concepts:
  Lagrangian mechanics, coupled oscillations
• Reasoning:
  We are asked to find the normal modes of coupled harmonic oscillators.
• Details of the calculation:
  (a)  T = ½M(dX/dt)² + ½m(d(X + x)/dt)² , U = ½kx².  L = T - U.
  L = ½(M+m)(dX/dt)² + ½m(dx/dt)² + m(dX/dt)(dx/dt) - ½kx².
  
  (b)  d/dt(∂L/∂(dX/dt)) -  ∂L/∂X = 0.   The coordinate X is cyclic.
  ∂L/∂(dX/dt)) = p_X =  constant.
  (M+m)(dX/dt) +  m(dx/dt) = constant,  (M+m)d²X/dt² =  -md²x/dt².
  
  d/dt(∂L/∂(dx/dt)) -  ∂L/∂x = 0.
  m(dx/dt)² + m(dX/dt) = -kx. 
  md²x/dt² - (m²/(M+m))d²x/dt² = -kx. 
  (Mm/(M+m))d²x/dt² = -kx.
  
  (c)  The normal mode frequencies are ω = 0 and ω² = (M + m)k/(Mm).
  
  Another approach:
  L = ½∑_ij[T_ij(dq_i/dt)(dq_j/dt) - k_ijq_iq_j].
  Here T₁₁ = (M+m), T₂₂ = m, T₁₂ = T₂₁ = m, k₁₁ = k₁₂ = k₂₁ = 0, k₂₂ = k.
  det(k_ij - ω²T_ij) = 0. 
  -ω²(M+m)(k - ω²m) - ω⁴m² = 0
  ω = 0 or ω² = (M + m)k/(Mm).
  The mode ω = 0 corresponds to all parts of the system moving with the same uniform velocity.
  For the mode ω = [(M + m)k/(Mm)^½ we have
  -[(M+m)(M + m)k/(Mm)]A_X + [-m(M + m)k/(Mm)] A_x = 0.
  A_X/A_x = -m/(M+m).  The center of mass of the system stays fixed.

Problem 4:

The potential energy of a body of mass m constrained to move in one dimension is U = kx⁴, where k is a constant.
(a)  What is the force on the body?
(b)  What is the Hamiltonian of this system?
(c)  The body moves from position x₁ at time t₁ to position x₂ at time t₂.  If these end points are fixed, the principle of stationary action requires which function to have an extremum for the x - t curve for this motion?

Solution:

• Concepts:
  The Hamiltonian of a system, the principle of least action
• Reasoning:
  F is a conservative force.  It is derivable from a potential.
  H = T + U.  H is a function of x and p.
• Details of the calculation:
  (a)  For a conservative force F = -dU/dx.   F = -4 kx³.
  (b)  The Hamiltonian H = T + U.  H is a function of x and p.
  H = p²/(2m) + kx⁴.
  (c)  Define the action S as the integral of the Lagrangian between two instances of time.  S = ∫_t1^t2L(q, dq/dt, t) dt.  δS = 0, the path taken by the system between times t₁ and t₂ and configurations q₁ and q₂ is the one for which the action is stationary (no change) to first order.
  Here S = ∫_t1^t2(½m(dx/dt)² - kx⁴) dt is required to have an extremum for the x - t curve for the motion,

Problem 5:

One end of a thin rod of length l, uniform density, and mass m is suspended from the ceiling of an elevator that is accelerating downward with constant acceleration a.  What is the period of small oscillations for this rod?

Solution:

• Concepts:
  A physical pendulum, motion in an accelerating frame
• Reasoning:
• Details of the calculation:
  For a physical pendulum subject to a constant downward force F = mg', the Lagrangian is given by
  L = ½I(dθ/dt)² + mg'l'cosθ, or
  L = ½I(dθ/dt)²- mg'l'θ²/2 to second order in the small quantities.
  Here l' is the distance from the suspension point to the CM and I is the moment of inertia about the suspension point.
  Equation of motion:  d²θ/dt² = -(mg'l'/I)θ.
  Period of small oscillations:  T = 2π/ω = 2π(I/(mg'l'))^½.
  In this problem l' = L/2 and I = ∫₀^L(m/L)x²dx = mL²/3
  The apparent weight F_g' = m_g' = m(g - a).  Since g and a both pint downward, g' = g - a.
  So T = 2π(2L²/(3(g - a)l)^½.

Problem 6:

(a)  Find the normalization constant A.
(b)  Find the expectation value of the 2 x 2 spin operator S_z.
(c)  Find the expectation value of the 2 x 2 spin operator S_x.
(d)  Find the expectation value of the 2 x 2 spin operator (S_y)².
(e)  What is the probability of finding +ћ/2 if S_z is measured?

Solution:

• Concepts:
  The mean value of an observable
• Reasoning:
  The expression for the mean value of an observable A in the normalized state |Ψ> is
  <A> = <Ψ|A|Ψ>.  If |Ψ> is not normalized then < A> = <Ψ|A|Ψ>/<Ψ|Ψ>.
  Here the system is the 2-dimensional state space of a spin-½ particle, and the observables are the Cartesian components of its spin.
• Details of the calculation:
  

  (a) Normalization: < Ψ|Ψ> = |A|² 

  2

  1-i

  	2
  	1+i

   = |A|²(4 + 2) = 1.

   

   

  Choose A = 1/√6 = positive, real.
   

  (b)  <S_z> = <χ|S_z|χ> = (ħ/12) 

  2

  1-i

  	1	0
  	0	-1

  	2
  	1+i

   = ħ/6.
   

   

  
  
  

  (c)  <S_x> = <χ|S_x|χ> = (ħ/12) 

  2

  1-i

  	0	1
  	1	0

  	2
  	1+i

   = ħ/3.

   

  
  

  (d)  (S_y)² = (ħ²/4) 

  	1	0
  	0	1

   = I*ħ²/4.

  
  
  
  (This is the operator S_y².)
  Therefore <S_y²> = (ħ²/4).
  (e)  |χ> = a |+> + b|-> = (2/√6 )|+> + ((1+i)/√6 )|->.
  P(+ћ/2) = |<+|χ>|² = 2/3.

Problem 7:

Reference frames S₁ and S₂ move with respect to reference frame S with speeds v₁ and v₂ respectively in the positive x-direction.  A stopwatch is at rest in S₁.  The time interval between the start and stop of events for the watch is t, as measured in S.  What are the corresponding time intervals as measured in S₁ and S₂?

Solution:

• Concepts:
  Relativistic kinematics, the Lorentz transformation
• Reasoning:
  We use the Lorentz transformation to find the space-time coordinates of two events in different reference frames.
• Details of the calculation:
  The time interval in S₁ is the proper time interval τ. 
  Δt = γ₁τ, or τ = Δt/γ₁ = Δt(1 - v₁²/c²)^½.
  We can use the Lorentz transformation from S to S₂ to find Δt₂.
  cΔt₂ = γ₂cΔt - γ₂β₂Δx. 
  But

  Δx = v₁Δt.  Therefore Δt₂ = γ₂(Δt - v₂v₁Δt/c²)  = γ₂Δt(1 - v₂v₁/c²).