# 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=Rsinωt

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:

### Need Graph Paper?

d=Rcosωt

d=Rsin (ωt+α)

d=Rcos (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.

Answer

Since *d* = *R* sin *ω**t*, 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:

### Example 2

The voltage of an alternating current circuit is given by

e=Ecos(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`.

Answer

We need to sketch:

e=Ecos(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

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"`.

Answer

Now, `f = 950 000`.

We need to find the **wavelength** to be able to draw 2
cycles.

We have: *b *= 2*πf* 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:

Graph of *e* = 0.014 cos(2π ×950000*t*). Units for *t* are 10^{-6} s

Easy to understand math videos:

MathTutorDVD.com

## 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=Asin(ωt+α).

Sketch 2 cycles of *D* as a function of *t* if

`A = 500\ "km"`, `ω = 3.60\ "rad/hr"` and ` α = 0`.

Answer

So we need to sketch: *D *= 500 sin(3.6*t*)

**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".

2. Using *e *= *E *cos(*ω**t* + *α*), sketch 2 cycles of the voltage as a
function of time if

E= 170 V,ω= 120πrad/s and `α = -π/3`.

Answer

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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