5.6. Modeling Harmonic Motion

Periodic behavior—behavior that repeats over and over again—is common in nature.

Let’s pull down the mass on the spring and then release it. Then, the mass start to oscillate. This is a simple example of harmonic motion.

When we relate the time with the height of the mass on the spring, we get a function. The graph of a function follows either the pattern of sine or cosine function.

A cycle is one complete vibration of an object, so the mass in the Figure completes one cycle of its motion between O and P.

Simple Harmonic Motion

If the equation describing the displacement $y$ of an object at time $t$ is either $y = a sin ω t$ or $y = a cos ω t, ω > 0,$ then the object is in simple harmonic motion. In this case,

Amplitude $= \| a \|$	Maximum displacement of the object
Period $= \frac{2 π}{ω}$	Time required to complete one cycle
Frequency $= \frac{ω}{2 π}$	Number of cycles per unit of time

Example

The displacement of a mass suspended by a spring is modeled by the function $y = 10 sin 4 π t$ , where $y$ is measured in inches and $t$ in seconds.

(a) Find the amplitude, period, and frequency of the motion of the mass.

(b) Sketch a graph of the displacement of the mass.

The figure illustrates the displacement of a mass suspended by a spring. Initially, the spring is in rest position. When the spring is stretched, the mass is displaced by y less than 0. When the spring is released, the mass is displaced by y greater than 0.

Solution

From the formulas for amplitude, period, and frequency we get

Amplitude $= | a | = 10$ in

Period $= \frac{2 π}{ω} = \frac{2 π}{4 π} = \frac{1}{2}$ s

Frequency $= \frac{ω}{2 π} = \frac{4 π}{2 π} = 2$ cycles per second (Hz)

By considering the given equation of the function $y = 10 sin 4 π t$ , amplitude and period, we get

Example

The displacement of a mass suspended by a spring is modeled by the function $y = 7 cos \frac{1}{4} t$ , where $y$ is measured in inches and $t$ in seconds.

(a) Find the amplitude, period, and frequency of the motion of the mass.

(b) Sketch a graph of the displacement of the mass.

Solution

From the formulas for amplitude, period, and frequency we get

Amplitude $= | a | = 7$ in

Period $= \frac{2 π}{ω} = \frac{2 π}{\frac{1}{4}} = 2 π \cdot \frac{4}{1} = 2 π \cdot 4 = 8 π$ s

Frequency $= \frac{ω}{2 π} = \frac{\frac{1}{4}}{2 π} = \frac{1}{4} \cdot \frac{1}{2 π} = \frac{1}{8 π}$ cycles per second (Hz)

By considering the given equation of the function $y = 7 cos \frac{1}{4} t$ , amplitude and period, we get

Example

Find a function that models the simple harmonic motion whose amplitude is 100 ft and period is 0.5 min. Assume that the displacement is at its maximum at time $t = 0 .$

Solution

Since the displacement is at its maximum at time $t = 0$ , the function is a cosine function $y = a c o s (ω t) .$

Since it has maximum at $t = 0$ and its amplitude is 100 ft, $a = 100 .$

Period: $0.5 = \frac{2 π}{ω}$

$0.5 ω = \frac{2 π}{ω} \cdot ω$

$0.5 ω = 2 π$

$ω = \frac{2 π}{0.5} = \frac{2}{0.5} \cdot π = 4 π$

By gathering all the information, we get $y = 100 cos (4 π t)$ .

Example

Find a function that models the simple harmonic motion whose amplitude is 1.7 m and frequency is 0.5 Hz. Assume that the displacement is zero at time $t = 0$ .

Solution

Since the displacement is zero at time $t = 0,$ the desired function is a sine function $y = a s i n (ω t)$ .

Since the amplitude is 1.7, $a = 1.7,$

Frequency( $= \frac{1}{p e r i o d}) :$ $\frac{ω}{2 π} = 0.5$

$\frac{ω}{2 π} \cdot 2 π = 0.5 \cdot 2 π$

$ω = π$

Therefore, the harmonic function is $y = 1.7 sin (π t) .$