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Marat540 [252]
2 years ago
11

The y-position of a damped oscillator as a function of time is shown in the figure.

Physics
1 answer:
NISA [10]2 years ago
4 0

(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

#SPJ1

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A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is
zvonat [6]
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5 0
3 years ago
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Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why d
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<u>Answer:</u>

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<u>Sources:</u>

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6 0
3 years ago
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms
inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) 6.81*10^5 N/c

b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

E_r_m_s = sqrt*(S / CE_o)

Where

C = 3.00 * 10^-^8 m/s;

E_o = 8.85*10^-^1^2 C^2/N.m^2;

S = 1.23*10^9 W/m^2

Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

E_r_m_s= 6.81 *10^5N/c

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

B_r_m_s = E_r_m_s / C;

= 6.81*10^5 N/c / 3*10^8m/s;

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