A projectile is shot upward with initial velocity and is subject both to gravity and to a drag force of magnitude , where c is constant.
1. Using direct integration techniques on Newton's second law, solve for v(x), the dependence of the particle's velocity on its height x.
  On its upward trajectory, both gravity and drag force act downward, giving
  
  We solve this by direct integration, noting as usual when F=F(v) that
  
  Integrating this forward for the motion from to height x, velocity v gives (being sloppy about dummy variables)
  
  or
  
  Multiplying both sides by -2c/m then exponentiating, we find
  
  Further manipulation lets us solve for v(x):
2. To what maximum height h does the projectile reach?
  Returning to our expression
  
  we reach our maximum height h when v=0:
3. From homework 1, assigned problem 2, we know that the projectile as it returns to earth obeys
  
  where the terminal velocity .
  With what velocity does the projectile strike the ground? (Express your answer in terms of and , the velocity the particle was initially shot up with in part (a).) Why does this make physical sense, from an energetics point of view?
  Note that our projectile climbs to height h given in part (b), then obeys the stated relation for v(x) as it falls. (This is the proper relation as, during the fall back to earth, the drag force acts upward leading to and v(x) as in homework 1, assigned problem 2.)
  The projectile falls from h as in part (b) to x=0 when it strikes the ground, with
  
  Plugging in our value for h gives
  
  Recognizing we rewrite this as
  
  or
  
  This makes sense physically from two points of view: first, as written in the first form, it's clear that the impact velocity v(x=0) must be less than , the terminal velocity at which drag balances gravity. Second, the impact velocity is also clearly less than , so energy is lost due to the always negative work done by the drag force. The projectile's initial kinetic energy is reduced by a factor upon its return to earth. The greater the initial velocity, the more energy is lost, as an everywhere stronger drag force acts over a greater distance 2h. A projectile shot up at half terminal velocity loses 20% of its energy to drag forces along its flight; one shot up at terminal velocity loses half its energy; while one shot up at twice terminal velocity loses 80%.
A particle of mass m moves in a conservative force field described by the potential energy

where a and c are positive constants.
1. What are the equilibrium positions for the particle?
2. Which are stable?
3. Give the frequency of oscillation about equilibrium, for any stable equilibrium points.
  We obtain the frequency of oscillation about equilibrium by identifying the quadratic term in a Taylor expansion of the potential about equilibrium
  
  with a harmonic oscillator potential , with . Thus
  
  At x = -a, . Thus
4. Sketch V(x) qualitatively.
  
  Note that V(0) = 0, and V(x) as approaches zero from values, respectively. The only extrema are the minimum at x=-a and the maximum at x=a.
5. For what range of total energy E does the particle undergo bound oscillations?
1. What is the equation of motion for the mass m?
2. What is the homogeneous solution?
3. What is the particular solution?
4. What is the motion x(t) of the mass m, if it begins at rest a distance beneath the table?
5. If we wait a very long time, what motion x(t) will we observe for the mass m?
Consider an LR circuit whose voltage source carries voltage (that is, B & O Figure 1-9 with C=0 and ).
1. Write the differential equation for the current i(t). What is the general solution for the homogeneous problem, for this ODE?
  For the LR circuit we have
  or in terms of current I(t) (using I instead of i to make time derivatives (dots) clear),
  
  The general solution for the homogeneous problem
  
  has where, plugging into the ODE,
  
  Thus .
2. Using the approach of B & O section 1-9, find a particular solution of the form
  
  That is, find and . Plot the phase lag .
  We replace our ODE by the complex ODE for complex I
  
  whose real part reproduces the original ODE , for real I. To solve this, we assume a solution of the form
  
  proportional to the source . Plugging this form into our ODE gives
  
  solved by
  
  To invert we write it in phasor form, where
  
  Plugging back into our solution,
  
  whose real part
  
  solves our original real ODE . Thus we have solution with
  
  The phase lag vanishes at and continuously increases until reaching as . It looks like
3. If the circuit initially has zero current, find the current i(t).
  Our general solution is the particular plus general homogeneous solution:
  
  At t=0 this has value
  
  which solves our initial condition I(0)=0 if
  
  Thus our solution is
  
  with and as in (b).
In analyzing LRC circuits, we introduce a complex impedance, which relates an applied sinusoidal voltage to the induced current in a generalization of Ohm's law:

Because Z is complex, it describes both the amplitude and phase lag of the response current.
1. Consider a circuit containing only a voltage source and a capacitor. By studying the particular solution to the Kirchoff's law ODE, derive the expression for capacitive impedance
  
  For each circuit, we consider the Kirchoff's law ODE for a series LRC circuit:
  
  For a capacitor only, this gives
  which we differentiate to find
  as . Thus
  which, by the definition , means that
2. Similarly, consider circuits containing first, only a voltage source and a resistor; and second, only a voltage source and an inductor. Treating Kirchoff's law as an ODE for the current i(t), derive expressions for the impedances and of a resistor and an inductor, respectively.
  For a resistor only, the Kirchoff's law ODE gives
  Thus, by the definition , Z = R. (Indeed, Z was defined to reduce to the resistance R in the case of a resistive circuit.)
  For an inductor only, the Kirchoff's law ODE gives
  Looking for particular solution gives
  thus by the definition ,
  (Note that we do have a particular solution of the form suggested, with .)
3. For a series LRC circuit, where the current through each circuit element is instantaneously the same, explain why your values for the impedances of R, L, and C agree with the conventional phasor diagram below, which shows voltage across a resistor being in phase with the current, while voltage across the inductor leads the current by a phase angle, and voltage across the capacitor lags the current by .
4. For the LR circuit of problem 4, note that is just the impedance Z of the circuit. Show that this impedance is just the sum .
  Our LR circuit of problem 4 had complex solution
  where
  Multiplying the first equation through by ,
  so is clearly playing the role of an impedance Z, relating the circuit's current I(t) to its total voltage across both LR elements. We found our solution had
5. For the LR circuit of problem 4, in the special case where , plot a phasor diagram like that of part (c), showing the phase relationships between the applied voltage , and the steady state: current I induced in the circuit, voltage across the resistor, and voltage across the inductor.