Damped simple harmonic motion question

  • Thread starter endusto
  • Start date
  • #1
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 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?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,914
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
6
0
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
6
0
can someone please tell me quickly if it does make sense that fd > fo? is this answer now correct?
 
  • #5
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,133
160
.
.
.
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)
.
.
.
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.
 

Related Threads on Damped simple harmonic motion question

  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
6K
Replies
8
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
732
Replies
1
Views
1K
  • Last Post
Replies
4
Views
400
  • Last Post
Replies
0
Views
4K
  • Last Post
Replies
3
Views
828
Top