# 5. Applications of Trigonometric Graphs

by M. Bourne Oscilloscope output - Filter modulation [Image source: ]

## 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)

### Example 1

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

"Period"=(2pi)/b=(2pi)/3.20=1.96

So the sketch will be:

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### Example 2

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:

### Example 3

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:

## Angular Velocity

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.

### Exercises

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".

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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.

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