5. Applications of Trigonometric Graphs
by M. Bourne
Oscilloscope output - Filter modulation [Image source: Mikael Altemark]
Simple Harmonic Motion
Any object moving with constant angular velocity or moving up and down with a regular motion can be described in terms of SIMPLE HARMONIC MOTION.
The displacement, d, of an object moving with SHM, is given by:
d = R sin ωt
where R is the radius of the rotating object and `ω` is the angular velocity of the object.
For an animation of this concept, go back to: sin animation.
NOTE: We may need to use one the following, depending on the situation:
d = R cos ωt
d = R sin (ωt + α)
d = R cos (ωt + α)
A point on a cam is `8.30\ "cm"` from the centre of rotation. Sketch 2 cycles of d as a function of t, given that d = 0 cm when t = 0 s and ω = 3.20 rad/s.
Since d = R sin ωt, we have
d = 8.30 sin 3.20t.
This sine curve will have amplitude `8.30` and period given by
So the sketch will be:
The voltage of an alternating current circuit is given by
e = E cos(ωt + α).
Sketch 2 cycles of the voltage as a function of time if
`E = 80\ "V"`, ` ω = 377\ "rad/s"` and `α = π/2`.
We need to sketch:
e = E cos(ωt + α)
e = 80 cos(377t + π/2)
This will have amplitude = `80`,
period `=(2pi)/377=0.0167` and
phase shift `= -c/b = -(π/2)/377 = -0.00417`
So our sketch is:
The signal received by a radio is given by
e = 0.014 cos(2πft),
where e is in volts and f is in Hz.
Draw 2 cycles of e for `f = 950\ "kHz"`.
Now, `f = 950 000`.
We need to find the wavelength to be able to draw 2 cycles.
We have: b = 2πf and we know that
`"Period" ` `=(2pi)/b=(2pi)/(2pixx950000) ` `=1.053xx10^(-6)`
So 2 wavelengths will be: `2xx1.053xx10^-6=2.105xx10^-6`
So, graphing for `0 < t < 2.105 xx 10^-6`, we have:
Another important result in this section is:
The angular velocity ω (in radians per second) of a rotating object, is given by:
ω = 2πf
where f is the frequency of the motion, in cycles per second.
1. A satellite is orbiting the earth so that its displacement D north of the equator is given by
D = A sin(ωt + α).
Sketch 2 cycles of D as a function of t if
`A = 500\ "km"`, `ω = 3.60\ "rad/hr"` and ` α = 0`.
So we need to sketch: D = 500 sin(3.6t)
Note: Be careful with units! If frequency is in cycles per minute, then the angular velocity will be in radians per minute. Also, be careful with "kilo".
Get the Daily Math Tweet!
IntMath on Twitter
2. Using e = E cos(ωt + α), sketch 2 cycles of the voltage as a function of time if
E = 170 V, ω = 120π rad/s and `α = -π/3`.
So `e = 170 cos (120πt - π/3)`
Amplitude is `170\ "V"`.`"Period"` ` = (2π)/b = (2π)/(120π)` ` =1/60` ` = 0.016667 `
`"Phase shift"` ` = -c/b = - (-π/3)/(120π)` ` = 1/360` ` = 0.0027778`
So our cosine curve will be shifted to the right by approximately `0.0028` seconds.