5.5 Modeling Harmonic Motion
5.5.1 Learning Objectives
By the end of this section, you should be able to:
- identify amplitude, period, frequency, and equilibrium;
- write models for simple harmonic motion;
- choose sine or cosine from initial conditions;
- interpret sinusoidal models in context; and
- describe damped harmonic motion.
5.5.2 Simple Harmonic Motion
Simple harmonic motion is periodic motion modelled by a sine or cosine function:
\[ d(t)=A\sin(\omega t) \quad\text{or}\quad d(t)=A\cos(\omega t), \]
where \(d(t)\) is displacement from equilibrium, \(|A|\) is amplitude, and \(\omega\) is angular frequency. The period and ordinary frequency are
\[ T=\frac{2\pi}{|\omega|}, \qquad f=\frac1T=\frac{|\omega|}{2\pi}. \]
Example 5.16 Reading a Motion Model
For \(d(t)=8\cos(3t)\), identify amplitude, period, and frequency.
Solution
\[ \boxed{\text{Amplitude }8,\quad T=2\pi/3,\quad f=3/(2\pi)} \]
Exercise 5.16 Find the amplitude and period of \(d(t)=-5\sin(\pi t/4)\).
Show answer
Answer: \(\boxed{\text{Amplitude }5,\quad \text{period }8}\)
5.5.3 Choosing Sine or Cosine
Cosine is convenient when motion begins at a maximum or minimum displacement. Sine is convenient when motion begins at equilibrium. A phase shift can represent any other starting position.
Example 5.17 Writing a Spring Model
A mass oscillates \(6\text{ cm}\) above and below equilibrium with period \(4\text{ s}\). At \(t=0\), it is at its maximum displacement. Write a model.
Solution
The amplitude is \(6\) and
\[ \omega=\frac{2\pi}{4}=\frac{\pi}{2}. \]
Beginning at a maximum suggests cosine:
\[ \boxed{d(t)=6\cos(\pi t/2)\text{ cm}}. \]
Exercise 5.17 Write a model with amplitude \(3\text{ m}\), period \(10\text{ s}\), beginning at equilibrium and moving upward.
Show answer
Answer: \(\boxed{d(t)=3\sin(\pi t/5)\text{ m}}\)
5.5.4 Sinusoidal Models with a Midline
Many periodic quantities oscillate around a nonzero average:
\[ y=A\sin(\omega(t-C))+D \]
or a corresponding cosine model. Here, \(D\) is the average or equilibrium value.
Example 5.18 Modeling a Ferris Wheel
A wheel has radius \(12\text{ m}\), centre \(14\text{ m}\) above the ground, and period \(40\text{ s}\). A rider begins at the lowest point. Write a height model.
Solution
The amplitude is \(12\), midline is \(14\), and \(\omega=2\pi/40=\pi/20\). Beginning at a minimum suggests negative cosine:
\[ \boxed{h(t)=14-12\cos(\pi t/20)\text{ m}}. \]
Exercise 5.18 For the Ferris-wheel model above, find the maximum and minimum heights.
Show answer
Answer: \(\boxed{\text{Minimum }2\text{ m};\quad \text{maximum }26\text{ m}}\)
5.5.5 Damped Harmonic Motion
In real systems, resistance often causes amplitude to decrease over time. A common damped harmonic motion model is
\[ d(t)=Ae^{-kt}\cos(\omega t), \qquad k>0. \]
The curves \(y=\pm|A|e^{-kt}\) form a shrinking envelope around the oscillation.
Example 5.19 Interpreting Damping
For \(d(t)=10e^{-0.2t}\cos(4t)\), find the initial amplitude and the amplitude after \(5\) seconds.
Solution
The initial amplitude is \(10\). After \(5\) seconds, the envelope has magnitude
\[ 10e^{-0.2(5)}=10e^{-1}\approx3.68. \]
\[ \boxed{\text{Initial amplitude }10;\quad \text{amplitude after }5\text{ s}\approx3.68} \]
Exercise 5.19 For \(d(t)=7e^{-0.1t}\sin(2t)\), find the amplitude envelope after \(10\) seconds.
Show answer
Answer: \(\boxed{7e^{-1}\approx2.58}\)
5.5.6 Conceptual Takeaways
- Harmonic motion repeats around an equilibrium position.
- Amplitude is the greatest displacement from equilibrium.
- Period and frequency are reciprocals.
- Initial conditions guide the choice of sine, cosine, and phase shift.
- Damping reduces amplitude without removing the underlying oscillation.
5.5.7 Skills You Should Be Able to Do
- Calculate amplitude, period, frequency, and angular frequency.
- Write simple harmonic motion models.
- Include a midline and phase shift when needed.
- Interpret periodic models with units.
- Analyze a damped amplitude envelope.
5.5.8 Practice Problems with Solutions
Find the amplitude and period of \(d(t)=4\cos(5t)\).
Show Solution
\[ \boxed{\text{Amplitude }4,\quad \text{period }2\pi/5}. \]
Find the frequency of \(d(t)=2\sin(6\pi t)\).
Show Solution
\[ f=\frac{6\pi}{2\pi}=3. \]
\[ \boxed{3\text{ cycles per unit of time}} \]
Write a cosine model with amplitude \(5\), period \(12\), and maximum displacement at \(t=0\).
Show Solution
Since \(\omega=2\pi/12=\pi/6\),
\[ \boxed{d(t)=5\cos(\pi t/6)}. \]
A buoy moves \(1.5\text{ m}\) above and below its average height with period \(8\text{ s}\). It begins at the average height moving upward. Write a displacement model.
Show Solution
Here \(A=1.5\) and \(\omega=2\pi/8=\pi/4\). Beginning at equilibrium moving upward suggests sine:
\[ \boxed{d(t)=1.5\sin(\pi t/4)\text{ m}}. \]
For \(h(t)=14-12\cos(\pi t/20)\), find the rider’s height after \(10\) seconds.
Show Solution
\[ h(10)=14-12\cos(\pi/2)=14. \]
\[ \boxed{14\text{ m}} \]
A damped oscillator is modelled by \(d(t)=12e^{-0.15t}\cos(3t)\). Find its amplitude envelope after \(8\) seconds.
Show Solution
\[ 12e^{-0.15(8)}=12e^{-1.2}\approx3.61. \]
\[ \boxed{3.61\text{ units, approximately}} \]