STATEMENT OF PROBLEM

When we first learned about trigonometric functions, we found that sin(t) went through 1 cycle in 2 Pi seconds. Furthermore, sin(2 Pi t) goes through 1 cycle per second and sin(2 Pi w t) goes through w cycles in 1 second -- as long as w is a constant. We call the constant, w, the frequency of oscillation. The following problem set is concerned with what happens if w is not a constant, i.e., w = w(t) is a function of time t. For example, suppose w(t) = t. If we graph y = sin(2 Pi w t) = sin(2 Pi t^2) for 5 seconds, it is clear that the frequency of oscillation increases with t.

Input := 

Plot[Sin[2 Pi t^2], {t, 0, 5}, AspectRatio -> Automatic,
 PlotPoints -> 50]

Output =

-Graphics-

1) Argue, however, that w(t) = t does NOT count the number of oscillations per second and is thus not the frequency of oscillation as follows:

a) Estimate the number, f, of cycles on [2, 3].

b) Note that your estimate of the frequency, f, does not satisfy w(2) <= f <= w(3). Explain why this implies that w(t) cannot be the frequency of oscillation.

2) In fact, if w(t) = t, then the frequency W(t), is W(t) = 2t. Show that, indeed, W(2) <= f <= W(3).

3) Argue that if w(t) = t^2, then the frequency is W(t) = 3t^2 by estimating the number of cycles sin(2 Pi w(t) t) has on 2 <= t <= 3 and comparing it with the values of W(t) on the same interval.

4) Argue further that if w(t) = 200 + (100/(2 Pi)) sin(2 Pi t) then the frequency is W(t) = 200 + 100t cos(2 Pi t) + (100/(2 Pi)) sin(2 Pi t) by estimating the number of cycles sin(2 Pi w(t) t) has on 1 <= t <= 1.04, multiplying that number by 25 to get the approximate number of cycles per second and comparing that value with the values of W(t) on the same interval.

5) Given w(t), how can you determine W(t)?

6) Suppose you are given W(t). How can you find w(t)?

7) Construct the function sin(2 Pi w(t) t) whose frequency W(t) increases linearly from 263 to 526 cycles per second over a period of 3 seconds. (A tight string oscillating at the given frequencies would play the middle C octave on a piano.)

8) Use Mathematica's Play[] command to play the function created in #7. You might be able to hear that the tones are weighted towards the top of the scale. This is due to the fact that the frequency is exponentially related to the octave:
F(n) = 263 2^n
is the frequency n octaves above middle-C. Find the frequency W(t) for t in [0, 3] and the corresponding function sin(2 Pi w(t)t) so that the tones increase at a constant rate over the 3 second interval.