(1) The period of the oscillator is 1 second.
(2) The damping coefficient is 0.93.
<h3>
What is period of oscillation?</h3>
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.
<h3>Damping coefficient</h3>
equation of the wave is given as;
y(t) = Ae^(-btx) cos(ωt)
<h3>at time, t = 0, y = 3.5</h3>
3.5 = Ae^(-0) cos(0)
3.5 = A x 1
A = 3.5 cm
<h3>at time, t = 1 cm, y = - 3cm</h3>
-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)
<h3>at time, t = 2 cm, y = - 2cm</h3>
-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: brainly.com/question/14058210
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