4. (a) All energy is in capacitor. (b) When
current is max. all energy is in inductor,
same amount of energy as part (a)
5. (a) One period is time to return to original
condition; cap discharges, charges oppositely,
discharges and charges again in one cycle.
(b) f = 1/T (c) see part (a)
7. (a) Refer back to chapter 14 for equations
on spring oscillators (c) Use same angular
frequency and LC oscillator equations.
9. Use resonance equation 35-19 but convert
to linear frequency.
12. Closing switch 1 creates RC circuit;
Closing switch 2 creates LR circuit and closing
switch 3 creates LC circuit. Write equations
for RC and LR time constants, solve for Cand
L, substitute into LC period equation.
24. (a) From resonant frequency equation,
small C gives higher f. So max. f is found
with min C and vice versa (6.0) (b) Find
new ratio of frequencies. Since new cap is
in parallel, its value will add to original
values and must equal new ratio. (36 pF)
26. When current is max, all energy is in
inductor, find this value. Conservation of
energy will give you the answer. (180 microcoulombs)
(b) Substitute equation for instantaneous
charge on cap into cap energy equation for
instantaneous energy eqn. Rate of change
will be derivative of this function. Use
trig identity of sin 2x = 2 sinx cos x to simplify in terms of sin only. Function
will be maximized when sin equals 1. Figure
out when in the cycle that will occur. (T/8)
(c) Maximize the derivative function and
put in terms of the period. Solve for the
period using values of L and C, plug into
maximized energy rate function and you have
the max rate. Easy, right? (66.7 W)
31. q/Q must equal .99 (see sample problem 35-4).
We must take into account the 50 cycles,
so our total time of oscillation is 50 times
the period of one oscillation. This makes
the exponent of e equal -RnT/2L, where n is the number of
oscillations. Assume the damped frequency
is nearly equal to the normal frequency.
37. This would be at the resonant frequency.