More Problems

Problem 1:

A mass m is attached to a horizontal spring.  Initially the mass is at rest.  At t = 0, a force F = F₀cosωt starts acting on the mass.  For t > 0, find the displacement x of the mass from its equilibrium position, assuming there is no friction and damping.  Assume that ω² ≠ k/m.

Solution:

• Concepts:
  Newton's second law, forced oscillations
• Reasoning:
  Equation of motion md²x/dt² = -kx + F₀cosωt
  We find a solution to the inhomogeneous equation d²x/dt² + (k/m)x = (F₀/m)cosωt
  and add the solution of the homogeneous equation d²x/dt² + (k/m)x = 0.
• Details of the calculation:
  d²x/dt² + ω₀x = (F₀/m)cosωt,  with ω₀² = k/m. 
  Try a solution x = Bcos(ωt).  This yields B = (F₀/m)/(ω₀² - ω²). 
  To find the most general solution we add the homogeneous solution x = Ccos(ω₀t + δ).
  Most general solution:  x(t) = Ccos(ω₀t + δ) + [(F₀/m)/(ω₀² - ω²)]cos(ωt).
  Initial conditions:  x(0) = dx/dt|_t=0 = 0.
  Ccos(δ) + [(F₀/m)/(ω₀² - ω²)] = 0,  ω₀Csin(δ) = 0  --> δ = 0,  C = -[(F₀/m)/(ω₀² - ω²)].
  x(t) = -[(F₀/m)/(ω₀² - ω²)]cos(ω₀t) + [(F₀/m)/(ω₀² - ω²)]cos(ωt).
  The motion is a superposition of two oscillations with different frequencies.  If |ω₀ - ω| << ω, we observe "beats".

Problem 2:

Assume the nucleus ^A_ZX decays.

Give the decay results if the nucleus decays via
(a)  alpha decay,
(b)  beta decay,
(c)  electron capture
(d)  gamma decay.

Solution:

• Concepts:
  Nuclear decay modes
• Reasoning:
  (a)  ^A_ZX  -->  ^A-4_Z-2X  + α
  (b)  β⁻ decay:  ^A_ZX  -->  ^A_Z+1X  + e^- + ν_e(bar)
  β⁺ decay:  ^A_ZX  -->  ^A_Z-1X  + e⁺ + ν_e
  (c)  ^A_ZX + e^-  -->  ^A_Z-1X + ν_e
  (d)  ^A_ZX*  --> ^A_ZX + γ

Problem 3:

An astronaut on the space station lets go of a flashlight whose mass is 1 kg and which is initially at rest with respect to the astronaut.  If the light output is 1 W and the battery lasts for 1 hour, estimate the final velocity of the flashlight relative to the astronaut.

Solution:

• Concepts:
  Photon momentum p = h/λ,  Newton/s third law
• Reasoning:
  Assume F = dp/dt is constant.  (This is a reasonable approximation for this relativistic rocket problem, since the final velocity will be very small.)
• Details of the calculation:
  dp/dt = 1 W/c.  p_final = (1J/s)(3600s)/3*10⁸ m/s) = 1.2*10^-5 Ns.
  v_final = 1.2*10^-5 m/s.

Problem 4:

The weak interactions (for example, beta decay) are mediated by massive particles called intermediate vector bosons, which are observed in accelerator experiments to have masses in the range mc² ~ (80 - 90)*10⁹ eV.  Assuming the weak interactions to occur because of the quantum-mechanical exchange of a virtual intermediate vector boson between two particles, estimate the maximum range of the weak force.

Solution:

• Concepts:
  The uncertainty principle ∆E∆t ~ ħ
• Reasoning:
  For the force carrier particle we have ∆E ~ mc² since the particle either exists or does not exist.  The virtual particle can propagate a distance no larger than R = c∆t in a time interval ∆t.  If we insert ∆t ~ ħ/∆E from above, we have  R ~ ħ/mc,
• Details of the calculation:
  R ~ ħ/(mc) = hc/(2πmc²) ~ (1240 eV nm)/(2π*10¹¹ eV)  ~ 2*10^-9 nm = 2*10^-18 m.

Problem 5:

A mirror creates an image of an object which is 3 times larger than the object.  The object is then moved to a new location, and the image is again 3 times larger than the object.
(a)  Is this a convex or concave mirror?
(b)  What is the ratio of the distance the object has been moved to the radius of curvature of the mirror?

Solution:

• Concepts:
  The mirror equation
• Reasoning:
  1/x_o + 1/x_i = 1/f,  M = -x_i/x_o.
• Details of the calculation:
  (a)  Only concave mirrors produce images larger than the object.  This is a concave mirror.
  (b)  One of the images is a real, inverted image and one of the images is a virtual, upright image.
  real image:  M = -3,  x_i = 3x_o.  x_o = 4f/3.
  virtual image:  M = 3, x_i = -3x_o.  x_o = 2f/3.
  Distance d the object has been moved:  d = 2f/3. 
  The radius of curvature R = 2f, d/R = 1/3.