# Damped simple harmonic motion question

## 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 im 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
Homework Helper
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

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

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

Redbelly98
Staff Emeritus
Homework Helper
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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.