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The y-position of a damped oscillator as a function of time is shown in the figure.
This function can be described by the y(t) = [tex]A_{0}[/tex][tex]e^{-btx}[/tex]cos(ωt) formula, where [tex]A_{0}[/tex] is the initial amplitude, b is the damping coefficient and ω is the angular frequency.
1. What is the period of the oscillator? Please, notice that the function goes through a grid intersection point.
2. Determine the damping coefficient.

The Yposition Of A Damped Oscillator As A Function Of Time Is Shown In The Figure This Function Can Be Described By The Yt TexA0textexebtxtexcosωt Formula Where class=

Sagot :

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(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

What is period of oscillation?

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

Damping coefficient

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

at time, t = 0, y = 3.5

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

at time, t = 1 cm, y = - 3cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

at time, t = 2 cm, y = - 2cm

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: https://brainly.com/question/14058210

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