Trigonometry
Notes on the Applications of Sine and Cosine Functions.
Simple Harmonic Motion: One of the basic applications of the sine and cosine functions is simple harmonic motion.
Simple Harmonic Motion
An object displays harmonic motion,
if and only if,
it moves in a sinusoidal wave and its
motion is modeled by a sine or cosine
function,
y = a sin(b t+c)+d
or
y = a cos(b t+c)+d
where y is the displacement from a point
of reference, usually the rest position, and
t is time.
The rest position is the place where the
object would naturally stay in place, but
since the object is actually in motion, it is
moving though its rest position.
Many objects have simple harmonic motion. Examples are:
Swinging Pendulums,
Bouncing Springs.
Actually, the above examples are only simple harmonic if there is no friction. With friction, the motion would be called damped
harmonic motion since the amplitude would decrease over time, and the object would come to rest at the rest position. But as long as
friction is negligible, we can use this function to model it.
The interpretation of the characteristics of the sinusoidal wave are:
Amplitude = A = the maximum distance from the rest position that the object moves.
Period = T = the time it takes the object to complete 1 cycle, a graphical period.
Phase Shift = the time difference from t = 0 to the start of the 1st cycle. The phase shift is usually not used unless we are comparing 2
objects in harmonic motion.
Vertical Shift = The distance from the rest position to the reference point, i.e. ground level. The vertical shift is also usually not used
since rest position is usually used as the reference point.
A related concept to the period is the frequency.
Frequency
The frequency is the number of cycles
per time unit. Thus, if the frequency = f,
1
1
1
b
f= =
=
=
T period 2π b 2π
and
2πf = b
The common units for frequency are
hertz.
1hertz = 1hz
cycle
= 1 second
So while the period is a measurement of the amount of time per cycle, the frequency is a measurement of the number of cycles per
time.
Sometimes the functions for simple harmonic motion are written as y = a sin(2πf t+c)+d and y = a cos(2πf t+c)+d to show the
frequency.
SCC:Rickman
Notes on the Applications of Sine and Cosine Functions.
Page #1 of 4
Example #1: A hanging spring is compressed 3in from its rest position and released at t = 0seconds. It returns to the same position
after 0.8 seconds. Find a) the amplitude of the motion, b) the period of the motion, c) the frequency of the motion, d) a function that
models the displacement, y, of the end of the spring from the rest position at time, t, and e) the displacement from the rest position at t
= 3min rounded to the tenths place.
►
a) Since the spring is compressed 3in., we can assume it will bounce from 3in above to 3in below the
rest position. Thus,
amplitude = A = 3in.
b) Because it takes 0.8 sec. to complete 1 cycle,
period = T = 0.8 seconds
c) Hence,
1
1
1
10
5
f=
= =
= Hz = Hz
period T 0.8s 8
4
d) Using the formulas from graphing,
2π
T=
b
2π 2π 20π 5π
b=
=
=
=
T 0.8
8
2
Also, since the displacement at t = 0sec. is a maximum, it would be better to use the cosine function. We could actually use either
sine or cosine, but to use sine would mean we would have to put in a phase shift to get a maximum at t = 0sec.
Therefore putting it together we get:
⎛ 5π ⎞
y = 3cos ⎜ t ⎟
⎝ 2 ⎠
⎛ 60sec ⎞
e) At t = 3min = 3min ⎜
⎟ = 180sec ,
⎝ 1min ⎠
⎛ 5π
⎞
y = 3cos ⎜ (180 ) ⎟ = 3cos ( 450π ) = -3in .
⎝ 2
⎠
Thus, the displacement is -3in. which would mean the end of the spring will be 3in below the rest
position. Therefore, it would be a minimum point.□
Example #2: A grandfather clock uses a pendulum to keep accurate time. This works because even though there is friction in the
mechanism, the frequency of the pendulum stays constant while the amplitude is decreasing, but let’s assume the friction is negligible
and thus the amplitude is constant. Each swing of the pendulum moves the second hand one position. Find a) the frequency of the
pendulum, b) the period, and c) a function that can be use to model the displacement of the end of the pendulum for an arbitrary
amplitude.
►
a) Because a second hand moves every second,
f = 1 cycle
= 1Hz .
sec
b) Then,
1
1
=
= 1second .
f 1Hz
c) Since the amplitude is arbitrary, we just leave it as A, and also because there was no mention as to what is happening at t = 0, we
have a lot of flexibility in finding a function. Thus, either of the functions below would work.
y = A sin ( 2π(1)t )
y = A cos ( 2π(1)t )
T=
y = A sin ( 2π t )
y = A cos ( 2π t )
□
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Notes on the Applications of Sine and Cosine Functions.
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Alternating Current: An AC current, like the ones in your house, has voltage that alternates from positive to neutral to negative to neutral and back to
positive. Neutral is the state when no current is flowing. Specifically, the voltage changes based on a sine or cosine function. Like
harmonic motion, either sine or cosine can be used so the choice is only one of convenience.
The interpretation of the characteristics of the sinusoidal wave are:
Amplitude = A = the maximum voltage from the average voltage, the center of the sinusoidal wave, usually neutral.
Period = T = the time it takes the object to complete 1 cycle, a graphical period.
Phase Shift = the time difference from t = 0 to the start of the 1st cycle. The phase shift is usually not used unless we are comparing 2
AC current.
Vertical Shift = the bias = The difference from the current’s average voltage to neutral.
Example #3: For a standard household AC circuit, find a) frequency, b) period , c) amplitude, and d) a function to model it.
►
a) Since a standard household AC current is 60Hz, f = 60Hz.
1
1
= second .
b) T =
60Hz 60
c) Here’s where things get a bit tricky. You might want to think that the 110V or 120V that a normal wall socket is rated is the
amplitude, but it’s not.
First, to keep from referring to 2 voltages let’s just use the average of 115V. Now the 115V is actually “115 V rms” or “115 volts
root-mean-square”. This is a type of average voltage. The actual maximum voltage is actually 2 times this. Therefore,
A = 115 2 V ≈ 162.63V
Meaning, the voltage oscillates from -163volts to +163volts.
d) Thus, using sine, V = 115 2 sin ( 2π(60) t ) = 115 2 sin (120π t ) . □
Example #4: 2 AC currents with the same maximum voltage and frequency, are combined to form another AC current. What would
be the net effect if the 2 currents are a) in phase, maximums occur at the same time, and b) half a period out of phase, 1 is at a
maximum when the other is at a minimum.
►
a) Since they have the same maximums at the same time, and likewise the same for the minimums, the voltage would double.
Vtotal = V1 + V2
= A sin ( 2πf t ) + A sin ( 2πf t )
= 2A sin ( 2πf t )
= 2V1 = 2V2
b) In this case, they would perfectly cancel each other out, and the voltage would be a constant 0V.
Vtotal = V1 + V2
⎡
1 ⎞⎤
⎛
= A sin ( 2πf t ) + A sin ⎢ 2πf ⎜ t − ⎟ ⎥
2f
⎝
⎠⎦
⎣
= A sin ( 2πf t ) + A sin [ 2πf t − π ]
1
1⎛1⎞ 1
period = ⎜ ⎟ =
2
2 ⎝ f ⎠ 2f
= A sin ( 2πf t ) − A sin ( 2πf t )
see figure to the right
sin ( θ − π ) = -sin ( θ )
=0
□
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Notes on the Applications of Sine and Cosine Functions.
Page #3 of 4
Sound: Pure tones like those used in many old video games have a waveform that is sinusoidal.
The interpretation of the characteristics of the sinusoidal wave are:
Amplitude = A = Volume.
Frequency = f = Pitch of the tone.
Phase Shift = the time difference from t = 0 to the start of the 1st cycle. The phase shift is usually not used
unless we are comparing 2 tones.
This one may be a little hard to see since sound is a compression wave, and it’s the pressure of the air that
is changing based on a sinusoidal function. See figure to the right.
Example #5: In music, going up 1 octave doubles the frequency of the tone. Middle C, C3, has a frequency of 261.63 Hz. What is the
frequency of the note a) C2, 1 octave below middle C, and b) C4, 1 octave above middle C.
►
f middle C 261.63Hz
a) f C2 =
=
= 130.815 Hz .
2
2
b) f C4 = 2f middle C = 2 ( 261.63Hz ) = 523.26 Hz .
Example #6: A tone generator is a device used to create different types of waveforms at different frequencies. If a tone generator is
set to generate a pure tune, it sending an AC current to the speaker. This current causes a magnet attached to the speaker’s cone to
move in simple harmonic motion. Finally, the movement of the cone creates a compression wave, the pure tone, that is heard. What
would be the relation of the amplitude and frequency of the sinusoidal functions for the voltage, displacement and pressure?
►
The frequency of all 3 would be the same, since the cone will move in time with the current, and the air will compress in time with
the cone’s movement.
The amplitudes would be for the most part proportional to each other, i.e. if the voltage doubled the volume will double, as long as
they didn’t too close to the limits of the equipment.□
Other waveforms can be generated by adding on these pure tones together. For example,
sin ( 2π t ) + 13 sin ( 2π(3) t ) + 15 sin ( 2π(5) t ) + 17 sin ( 2π(7) t ) + 19 sin ( 2π(9) t ) + ...
= a pulse-wave, the waveform of a piano.
This is the basics of how older synthesizers and computers generated different sounds.
SCC:Rickman
Notes on the Applications of Sine and Cosine Functions.
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