If one stereo system is capable of producing 20 watts of sound power
and another can put out 50 watts, how many times greater is the amplitude
of the sound wave that can be created by the more powerful system?
(Assume they are playing the same music.)
Many fish have an organ known as a swim bladder, an air-filled cavity
whose main purpose is to control the fish's buoyancy an allow it to keep
from rising or sinking without having to use its muscles. In some fish,
however, the swim bladder (or a small extension of it) is linked to the ear
and serves the additional purpose of amplifying sound waves. For a typical
fish having such an anatomy, the bladder has a resonant frequency of 300
Hz, the bladder's Q is 3, and the maximum amplification is about a factor
of 100 in energy. Over what range of frequencies would the amplification
be at least a factor of 50?
As noted in section 2.4, it is only approximately true that the amplitude
has its maximum at
Being more careful, we should
actually define two different symbols,
and fres for the slightly
different frequency at which the amplitude is a maximum, i.e. the actual
resonant frequency. In this notation, the amplitude as a function of
Show that the maximum occurs not at fo but rather at the frequency
Hint: Finding the frequency that minimizes the quantity inside the square
root is equivalent to, but much easier than, finding the frequency that
maximizes the amplitude.
(a) Let W be the amount of work done by friction per cycle of oscillation,
i.e. the amount of energy lost to heat. Find the fraction of the
original energy E that remains in the oscillations after n cycles of motion.
(b) From this prove the equation (1 - W/E)Q = e - 2π (recalling that the
number 535 in the definition of Q is e2π).
(c) Use this to prove the approximation 1/Q≈ (1/2π)W/E. [Hint: Use the
approximation ln(1+x)≈ x, which is valid for small values of x.]
The goal of this problem is to refine the proportionality FWHM fres/Q into the equation FWHM=fres/Q, i.e. to prove that the constant of
proportionality equals 1.
(a) Show that the work done by a damping force F=-bv over one cycle of
steady-state motion equals Wdamp=-2π2bfA2. Hint: It is less confusing to
calculate the work done over half a cycle, from x=-A to x=+A, and then
(b) Show that the fraction of the undriven oscillator's energy lost to
damping over one cycle is |Wdamp| / E = 4π2bf / k.
(c) Use the previous result, combined with the result of problem 4, to
prove that Q equals k/2πbf .
(d) Combine the preceding result for Q with the equation FWHM=b/2πm
from section 2.4 to prove the equation FWHM=fres/Q.