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Simple harmonic motion

A type of motion described as simple harmonic motion    involves a restoring force but assumes that the motion will continue forever. Imagine a weighted object hanging on a spring, When that object is not disturbed, we say that the object is at rest, or in equilibrium. If the object is pulled down and then released, the force of the spring pulls the object back toward equilibrium and harmonic motion begins. The restoring force is directly proportional to the displacement of the object from its equilibrium point. When t = 0 , d = 0.

Simple harmonic motion

We see that simple harmonic motion    equations are given in terms of displacement:

d = a cos ( ω t )   or   d = a sin ( ω t )

where | a | is the amplitude, 2 π ω is the period, and ω 2 π is the frequency, or the number of cycles per unit of time.

Finding the displacement, period, and frequency, and graphing a function

For the given functions,

  1. Find the maximum displacement of an object.
  2. Find the period or the time required for one vibration.
  3. Find the frequency.
  4. Sketch the graph.
    1. y = 5 sin ( 3 t )
    2. y = 6 cos ( π t )
    3. y = 5 cos ( π 2 t )
  1. y = 5 sin ( 3 t )
    1. The maximum displacement is equal to the amplitude, | a | , which is 5.
    2. The period is 2 π ω = 2 π 3 .
    3. The frequency is given as ω 2 π = 3 2 π .
    4. See [link] . The graph indicates the five key points.
      Graph of the function y=5sin(3t) from 0 to 2pi/3. The five key points are (0,0), (pi/6, 5), (pi/3, 0), (pi/2, -5), (2pi/3, 0).
  2. y = 6 cos ( π t )
    1. The maximum displacement is 6.
    2. The period is 2 π ω = 2 π π = 2.
    3. The frequency is ω 2 π = π 2 π = 1 2 .
    4. See [link] .
      Graph of the function y=6cos(pi t) from 0 to 3.
  3. y = 5 cos ( π 2 ) t
    1. The maximum displacement is 5.
    2. The period is 2 π ω = 2 π π 2 = 4.
    3. The frequency is 1 4 .
    4. See [link] .
      Graph of the function y=5cos(pi/2 t) from 0 to 4.
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Damped harmonic motion

In reality, a pendulum does not swing back and forth forever, nor does an object on a spring bounce up and down forever. Eventually, the pendulum stops swinging and the object stops bouncing and both return to equilibrium. Periodic motion in which an energy-dissipating force, or damping factor, acts is known as damped harmonic motion    . Friction is typically the damping factor.

In physics, various formulas are used to account for the damping factor on the moving object. Some of these are calculus-based formulas that involve derivatives. For our purposes, we will use formulas for basic damped harmonic motion models.

Damped harmonic motion

In damped harmonic motion    , the displacement of an oscillating object from its rest position at time t is given as

f ( t ) = a e c t sin ( ω t ) or   f ( t ) = a e c t cos ( ω t )

where c is a damping factor, | a | is the initial displacement and 2 π ω is the period.

Modeling damped harmonic motion

Model the equations that fit the two scenarios and use a graphing utility to graph the functions: Two mass-spring systems exhibit damped harmonic motion at a frequency of 0.5 cycles per second. Both have an initial displacement of 10 cm. The first has a damping factor of 0.5 and the second has a damping factor of 0.1.

At time t = 0 , the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.

We are given the frequency f = ω 2 π of 0.5 cycles per second. Thus,

   ω 2 π = 0.5      ω = ( 0.5 ) 2 π         = π

The first spring system has a damping factor of c = 0.5. Following the general model for damped harmonic motion, we have

f ( t ) = 10 e 0.5 t cos ( π t )

[link] models the motion of the first spring system.

Graph of the first spring system, f(t) = 10(e^(-.5t))cos(pi*t), which begins with a high amplitude and quickly decreases.

The second spring system has a damping factor of c = 0.1 and can be modeled as

f ( t ) = 10 e 0.1 t cos ( π t )

[link] models the motion of the second spring system.

Graph of f(t) = 10(e^(-.1t))cos(pi*t), which begins with a high amplitude and slowly decreases (but has a high frequency).
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Questions & Answers

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Joe Reply
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a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
Divya Reply
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
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100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
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Practice Key Terms 2

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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