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Finding a cosine function that models damped harmonic motion

Find and graph a function of the form y = a e c t cos ( ω t ) that models the information given.

  1. a = 20 , c = 0.05 , p = 4
  2. a = 2 , c = 1.5 , f = 3

Substitute the given values into the model. Recall that period is 2 π ω and frequency is ω 2 π .

  1. y = 20 e 0.05 t cos ( π 2 t ) . See [link] .
    Graph of f(t) = 20(e^(-.05t))cos(pi/2 * t), which begins with a high amplitude and slowly decreases.
  2. y = 2 e 1.5 t cos ( 6 π t ) . See [link] .
    Graph of f(t) = 2(e^(-1.5t))cos(6pi * t), which begins with a small amplitude and quickly decreases to almost a straight line.
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The following equation represents a damped harmonic motion model:   f ( t ) = 5 e 6 t cos ( 4 t ) Find the initial displacement, the damping constant, and the frequency.

initial displacement =6, damping constant = -6, frequency = 2 π

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Finding a sine function that models damped harmonic motion

Find and graph a function of the form y = a e c t sin ( ω t ) that models the information given.

  1. a = 7 , c = 10 , p = π 6
  2. a = 0.3 , c = 0.2 , f = 20

Calculate the value of ω and substitute the known values into the model.

  1. As period is 2 π ω , we have
        π 6 = 2 π ω   ω π = 6 ( 2 π )     ω = 12

    The damping factor is given as 10 and the amplitude is 7. Thus, the model is y = 7 e 10 t sin ( 12 t ) . See [link] .

    Graph of f(t) = 7(e^(-10t))sin(12t), which spikes close to t=0 and quickly becomes almost a straight line.
  2. As frequency is ω 2 π , we have
       20 = ω 2 π 40 π = ω

    The damping factor is given as 0.2 and the amplitude is 0.3. The model is y = 0.3 e 0.2 t sin ( 40 π t ) . See [link] .

    Graph of f(t) = .3e^(-.2t)sin(40pi*t), which has a small amplitude but quickly decreases to the appearance of a straight line. The frequency is so high that, in this scaling, the function looks like a solid shape. The zoom in cut out of the graph shows the actual sinusoidal image of the function.
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Write the equation for damped harmonic motion given a = 10 , c = 0.5 , and p = 2.

y = 10 e 0.5 t cos ( π t )

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Modeling the oscillation of a spring

A spring measuring 10 inches in natural length is compressed by 5 inches and released. It oscillates once every 3 seconds, and its amplitude decreases by 30% every second. Find an equation that models the position of the spring t seconds after being released.

The amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we will write A as a negative value. We can write the amplitude portion of the function as

A ( t ) = 5 ( 1 0.30 ) t

We put ( 1 0.30 ) t in the form e c t as follows:

0.7 = e c     c = ln .7     c = 0.357

Now let’s address the period. The spring cycles through its positions every 3 seconds, this is the period, and we can use the formula to find omega.

3 = 2 π ω ω = 2 π 3

The natural length of 10 inches is the midline. We will use the cosine function, since the spring starts out at its maximum displacement. This portion of the equation is represented as

y = cos ( 2 π 3 t ) + 10

Finally, we put both functions together. Our the model for the position of the spring at t seconds is given as

y = 5 e 0.357 t cos ( 2 π 3 t ) + 10

See the graph in [link] .

Graph of the function y = -5e^(-.35t)cos(2pi/3 t) + 10 from 0 to 24. It starts out as waves with a high amplitude and decreases to almost a straight line very quickly.
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A mass suspended from a spring is raised a distance of 5 cm above its resting position. The mass is released at time t = 0 and allowed to oscillate. After 1 3 second, it is observed that the mass returns to its highest position. Find a function to model this motion relative to its initial resting position.

y = 5 cos ( 6 π t )

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Finding the value of the damping constant c According to the given criteria

A guitar string is plucked and vibrates in damped harmonic motion. The string is pulled and displaced 2 cm from its resting position. After 3 seconds, the displacement of the string measures 1 cm. Find the damping constant.

The displacement factor represents the amplitude and is determined by the coefficient a e c t in the model for damped harmonic motion. The damping constant is included in the term e c t . It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation

a e c ( t + 3 ) = 1 2 a e c t

Use algebra and the laws of exponents to solve for c .

  a e c ( t + 3 ) = 1 2 a e c t e c t e 3 c = 1 2 e c t Divide out  a . e 3 c = 1 2   Divide out  e c t . e 3 c = 2 Take reciprocals .

Then use the laws of logarithms.

e 3 c = 2   3 c = ln 2     c = ln 2 3

The damping constant is ln 2 3 .

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