PROBLEMS FROM FISHBANE and other ADDITIONAL PROBLEMS
JULY 31 - AUG 6
Problems from Fishbane, Gasiorowicz, and Thornton (2nd edition)
CHAPTER 28
57. Find a differential equation which can be solved for Q(t), the
charge as a function of time, on a discharging capacitor. The
figure for this problem is Fig. 28-13, with an initial charge
of Q_0 on capacitor C when the switch is moved to position b at
t = 0. Let positive current represent a charging capacitor,
i.e. i = (dQ/dt). Show by direct substitution that Q(t) =
(Q_0)e^(-t/RC) is a solution of this differential equation.
CHAPTER 33
46. The figure for this problem is Fig. 31-17. After being at
position a for a long time, switch S is thrown to b. Find the
differential equation which can be solved for the resulting
I(t), and show that I(t) = (EMF/R)e^(-tR/L) is a solution of
that differential equation.
EXTRA PROBLEMS FOR CHAPTER 33
G1 and G2:
The differential equation for an undriven LRC circuit is
(q/C) + R(dq/dt) + L(d^2q/dt^2) = 0 Eq (1)
Your task is to show that
q(t) = (Q_0)e^(-(alpha)t)cos((omega)t)
is a solution of this differential equation, where Q_0, alpha,
and omega are constants.
G1. Substitution of q(t) in Eq (1) gives terms in sin((omega)t)
and terms in cos((omega)t). Show that the sin((omega)t) terms
sum to zero only if alpha = R/2L.
G2. Show that the cos((omega)t) terms sum to zero only if
omega^2 = omega0^2 - alpha^2 where omega0^2 = 1/LC.
G3 and G4:
The differential equation for a driven LRC circuit is
(q/C) + R(dq/dt) + L(d^2q/dt^2) = (EMFm)sin((omegaD)t) Eq (2)
where (EMFm)sin((omegaD)t) is the driving EMF as a function of time,
EMFm is the amplitude of that driving EMF, and omegaD is the driving
angular frequency. Your task is to show that
q(t) = -(Q_0)cos((omegaD)t - phi)
is a solution of this differential equation, where Q_0 and phi are
constants.
G3. Substitution of q(t) in Eq (2) gives terms in sin((omegaD)t)
and terms in cos((omegaD)t). HINT: Use the trig identities for
sin(alpha+-beta) and cos(alpha+-beta) as found on page A9 of HRW.
Show that the sum of the cos((omegaD)t) terms = 0 only if
tan(phi) = (1/R)(X_L-X_C)
where X_L and X_C are the inductive and capacitive reactances.
G4. (a) Show that the sum of the sin((omegaD)t) terms =
(EMFm)sin((omegaD)t) only if
EMFm/(omegaD)Q_0 = R/cos(phi)
and therefore that Z = R/cos(phi), where Z is the inductance.
HINT: Use sin^2(phi) + cos^2(phi) = 1.
(b) From (a), show that Z = sqrt((X_L-X_C)^2+R^2).
HINT: sec^2(phi) = 1 + tan^2(phi).