Changing Times at Cuban Universities: Looking into the Transition towards a Social, Entrepreneurial and
4. Findings
12.1 PERIODIC FUNCTIONS
Exercises
12.1.1. Use appropriate trigonometric identities to show that
cos ωt + cos(ωt + θ) = 2 cos(θ/2) cos(ωt + θ/2) .
Solution:
Expand cos(ωt + θ) and cos(ωt + θ/2). The equation to be proved then takes the form
cos ωt + cos ωt cos θ− sin ωt sin θ =
2 cos(θ/2) cos ωt cos(θ/2)− 2 cos(θ/2) sin ωt sin(θ/2) . In the first term of the right-hand side, replace 2 cos2(θ/2) by 1 + cos θ; in the second right-hand term, replace 2 cos(θ/2) sin(θ/2) by sin θ. The two sides of the equation are then identical in form.
12.1.2. Plot sin(1.5x) over a range of x sufficient to identify the smallest interval on which this func-tion is periodic.
Solution:
Execute one of
> plot(sin(1.5*x),x=0 .. 4*Pi/3);
Plot[Sin[1.5*x],{x,0,4*Pi/3}]
The function sin(1.5x) is periodic with period 4π/3. Here is its graph.
190
12.1. PERIODIC FUNCTIONS 191 12.1.3. (a) Given f (x) = sin(πx/2) defined on the range (0, 2), assume that its extension to x values beyond that range is periodic (with wavelength 2). Sketch or plot f (x) and its periodic extension for the range (−3, 4).
(b) For the same f (x), assume it to be defined on the range (−1, 1) and sketch or plot it and its periodic extension for the range (−3, 4).
(c) If your answers for parts (a) and (b) differ, explain briefly why.
Solution:
(a) One can plot f (x) for (0, 2) and sketch its repetition by hand to span the range (−3, 4). To do this with symbolic computing requires code such as, in maple,
> Psin:= proc(xx) local x;
> x := xx; while (x>2) do x:=x-2 end do;
> while (x<0) do x:=x+2 end do;
> sin(Pi*x/2); end proc:
> plot(’Psin(x)’, x=-3 .. 4);
The plot command is confused by the call to Psin if its evaluation is not delayed by adding the single quotes.
In mathematica, the corresponding code is PSin[xx_] := Module[{x}, x = xx;
While[x > 2, x = x-2]; While[x < 0, x = x+2];
Sin[Pi*xx/2] ]
Plot[PSin[x], {x, -3, 4}]
(b) To make plots for the periodic extension of f (x) from the range (−1, 1), in the above coding change x > 2 to x > 1 and x < 0 to x < -1.
The plots for parts (a) and (b) are shown here:
(c) The plots differ because f (x) is not periodic in an interval of length 2.
12.1.4. A wave distribution is described by f (x, t) = 4 cos(2x + 3t− 0.4). As perceived by an observer at x = 0.5,
(a) Determine the velocity with which the distribution passes the observer, and state the direction in which it is moving.
(b) Find the wavelength, frequency, angular frequency, and period of the distribution.
(c) Find the times at which the observer will see (i) a (positive) maximum amplitude, and (ii) a node.
192 CHAPTER 12. FOURIER SERIES Solution:
Start by finding the period, which is the time required (at constant x) for 2x + 3t− 0.4 to increase by 2π, i.e., τ = 2π/3. The frequency is then ν = 1/τ = 3/2π and the angular frequency is ω = 2πν = 3. The wavelength is the change in x (at constant t) required for 2x + 3t− 0.4 to increase by 2π, i.e., λ = π. The velocity is v = λν = 3/2. Thus,
(a) The velocity is 3/2; a point of constant f moves to smaller x as t increases;
the distribution is moving in the−x direction.
(b) λ = π, ν = 3/2π, ω = 3, τ = 2π/3.
(c) (i) f (x, t) is a maximum when 2x + 3t− 0.4 = 2nπ, with n an integer. For x = 0.5, this condition is 3t + 0.6 = 2nπ, or t =−0.2 + 2nπ/3.
(ii) Nodes occur when 2x + 3t− 0.4 = (n + 12)π, with n an integer. For x = 0.5, this condition is 3t + 0.6 = (n +12)π, or t =−0.2 + (n +12)π/3.
12.1.5. A sinusoidal wave distribution is moving in the direction of the vector ˆex+ ˆey at velocity 250 m/s and with angular frequency 1250 s−1. The maximum amplitude of the wave is 3.0 mm. Determine the wavelength and period of the oscillation, and write an equation for its amplitude as a function of position and time.
Solution:
From ω = 1250 s−1, ν = 1250/2π s−1, and τ = 2π/1250 s. From λν = v, we find λ = 250/(1250/2π) = 2π/5 m.
The amplitude of the wave distribution must be a function of x + y, and a unit change in x + y corresponds to a displacement of the distribution by a distance 1/√
2. The wave distribution will therefore have an amplitude given by ψ(x, y, t) = A sin
[(2π λ
) (x + y
√2 )
− ωt + δ ]
Here δ is an arbitrary phase. The term ωt enters with a minus sign because the wave is to move in the positive direction of ˆex+ ˆey. Inserting the data for the current problem,
ψ(x, y, t) = 0.003 sin ( 5
√2(x + y)− 1250 t + δ )
m .
12.1.6. The fundamental frequency of a string of length 1 m clamped at both ends is 50 s−1. (This is the standing-wave oscillation of longest wavelength.) On a long string maintained under similar conditions (string density and tension), how long would it take a traveling wave to move 100 m?
Solution:
The wavelength of the fundamental frequency is two times the length of the string: λ = 2 m. The velocity is then v = λν = 2× 50 = 100 m/s. Travel of 100 m would take 1 s.
12.1.7. A piston is caused to move back and forth horizontally by being attached via a long rod to a point on the circumference of a rotating wheel) see Fig. 12.4. Find (in seconds) the period of oscillation of the piston when the wheel is rotating at 3000 rpm (revolutions per minute).
12.1. PERIODIC FUNCTIONS 193
Figure 12.4: Piston for Exercise 12.1.7.
Solution:
The period of the rotating wheel is the same as that of the oscillation of the piston. The wheel has frequency 3000/60 = 50 s−1 and therefore period 1/50 s.
12.1.8. If two waves of the same wavelength and of unit intensity (in arbitrary units) and traveling in the same direction are superposed, determine the intensity of the combined wave distribution:
(a) If the two waves are in phase, (b) If they differ in phase by 30◦, (c) If they differ in phase by 90◦, (d) If they differ in phase by 180◦.
Solution:
The intensity is proportional to the square of the amplitude.
(a) The net amplitude is twice that of either wave, so the intensity is increased by a factor 22, to 4 units.
(b) Using Eq. (12.7), the combined wave will (noting that 30◦ = π/6) have amplitude 2 cos(π/12), and therefore intensity 4 cos2(π/12) = 2 +√
3.
(c) Again using Eq. (12.7) and noting that 90◦= π/2, the combined wave has amplitude 2 cos(π/4) and intensity 4 cos2(π/4) = 2.
(d) The two waves are (at all times) of the same magnitude but opposite in sign; the resultant intensity is zero.
12.1.9. A wave of frequency 1 megahertz (106s−1) is multiplied by a wave of frequency 100 s−1. It is sometimes said that the low-frequency signal is used to modulate (more precisely, “amplitude-modulate”) the high-frequency signal. What frequencies will be present in the combined fre-quency distribution?
Solution:
We need to identify the modulated wave as a sum of contributions. For sim-plicity, assume that the phases of the two signals correspond to the product cos ωt cos αt. We then have
cos ωt cos αt =1 2 [
cos(ω + α)t + cos(ω− α)t] .
In the current problem, ω = 2π×106and α = 2π×100, so ω±α = 2π(106±100).
These factors correspond to frequencies 106± 100.