Damped simple harmonic motion question

So your answer is correct.In summary, the undamped frequency f0 of a damped oscillator with a frequency fd of 100 Hz and a ratio of one half between the amplitudes of two successive maxima is approximately 99.4 Hz. The equation for the position of the oscillator is x(t) = e^-(yt) * A cos (wd + ϕ), where y is a constant that affects how quickly the oscillator is damped, wd is the damped angular velocity, wo is the undamped angular velocity, and ϕ is an initial angle.
  • #1
endusto
6
0

Homework Statement



The frequency fd of a damped oscillator is 100 Hz, and the ratio of the amplitudes of two successive maxima is one half. What is the undamped frequency f0 of this oscillator?

Homework Equations


this is the equation in my textbook for the position at time t of an underdamped harmonic oscillator:
x(t) = e^-(yt) * A cos (wd + ϕ)

where y (really supposed to be gamma) is a constant that affects how quickly the oscillator is damped and w (really supposed to be omega) is the angular velocity and ϕ is just an initial angle

i chose the underdamped equation because i believe there can not be any maxima if it is overdamped or critically damped.

Td = 2pi / wd = 2pi / sqrt(Wo^2 - y^2) = 0.01 (Td is the damped period, wd is damped angular velocity, wo is undamped angular velocity)

and the only other equations i used are T = 2π/w and T = 1/f

where w is once again the angular velocity

The Attempt at a Solution


one maxima occurs at t=0 and the next one at t=T. first i will calculate wd for the position equations.

Td = 1/fd = 1/100 = 0.01

Td = 2pi/w
0.01 = 2pi/wd
wd = 2pi/0.01 = 200pi

x(t) = e^-(yt) * A* cos(wd * t)

x(0) = e^-(y * 0) * A * cos(200pi * 0)) = A

so the first maxima is just an amplitude, which makes sense. i will now find the height of the next maxima, which occurs when t = T

x(T) = x(0.01) = e^-(0.01y) * (Acos(200pi)) = A*e^-(0.01y)

x(T) = A*e^-(0.01y) ( because cos(200pi) = 1)

x(T) makes sense as a maxima because its just a damped amplitude, there is no cos factor making it smaller. now the question said "the ratio of the amplitudes of two successive maxima is one half" so...

x(T)/x(0) = A*e^-(0.01y) / A = 1/2

e^-(0.01y) = 1/2

-0.01y = ln (1/2)
y = -100 ln (1/2) = 69.3147181

now I will solve for Wo, which will give To, which will give Fo. I believe my problem is somewhere in these steps (unless I am using some completely wrong equations...)

Td = 2pi / wd = 2pi / sqrt(Wo^2 - y^2) = 0.01
0.01 = 2pi / sqrt(wo^2 + 69.3147181^2)
sqrt(wo^2 + 69.3147181^2) = 200pi
wo^2 + 69.3147181^2 = sqrt(200pi)
wo = sqrt(sqrt(200pi) + 69.3147181) = 69.4952979

so wo is slightly greater than wd, which i expect

To = 2pi/wo = 0.0904116609
fo = 1/To = 11.0605202

now this is what i don't get. why is fo LESS than fd? did i do something wrong?
 
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  • #2
endusto said:
sqrt(wo^2 + 69.3147181^2) = 200pi
wo^2 + 69.3147181^2 = sqrt(200pi)

It should be

wo^2 + 69.3147181^2 = [(200pi)]^2.


ehild
 
  • #3
thanks ehild. so now i have

wo^2 + 69.3147181^2 = 200pi^2
wo = sqrt((200pi)^2 - 69.3147181^2) = 624.483503

To = 2pi/wo = 0.0100614112
Fo = 1/To = 99.3896363

very good makes sense thanks a lot
 
  • #4
can someone please tell me quickly if it does make sense that fd > fo? is this answer now correct?
 
  • #5
endusto said:
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.
.
Td = 2pi / wd = 2pi / sqrt(Wo^2 - y^2) = 0.01 (Td is the damped period, wd is damped angular velocity, wo is undamped angular velocity)
.
.
.
Td = 2pi / wd = 2pi / sqrt(Wo^2 - y^2) = 0.01
0.01 = 2pi / sqrt(wo^2 + 69.3147181^2)
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Looks like you switched a +/- sign in your work. The undamped frequency should be higher than the damped frequency; damping will slow down the oscillator.
 

1. What is damped simple harmonic motion?

Damped simple harmonic motion refers to the motion of an object that oscillates back and forth around an equilibrium point, while also experiencing a damping force that reduces its amplitude over time.

2. How is damped simple harmonic motion different from regular simple harmonic motion?

In regular simple harmonic motion, there is no damping force acting on the object and the amplitude remains constant. In damped simple harmonic motion, the damping force causes the amplitude to decrease over time.

3. What causes damping in simple harmonic motion?

Damping in simple harmonic motion can be caused by various factors such as air resistance, friction, and internal friction within the object itself.

4. How is the damping factor related to the amplitude in damped simple harmonic motion?

The damping factor, also known as the coefficient of damping, determines how quickly the amplitude decreases in damped simple harmonic motion. A higher damping factor means a faster decrease in amplitude.

5. Can damped simple harmonic motion ever reach a state of equilibrium?

Yes, damped simple harmonic motion can reach a state of equilibrium where the amplitude no longer decreases and the object stops oscillating. This can happen when the damping force is equal to the restoring force of the system.

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