Damped simple harmonic motion question

[endusto]

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?

 

[ehild]

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

 

[endusto]

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

 

[endusto]

can someone please tell me quickly if it does make sense that fd > fo? is this answer now correct?

 

[Redbelly98]

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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)
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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.