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Driven Damped Harmonic Oscillator Problem

  • Thread starter Walczyk
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5
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A mass m moves along the x-axis subject to an attractive force given by [tex]\frac {17} {2} \beta^2 m x [/tex] and a retarding force given by [tex]3 \beta m \dot{x}[/tex], where [tex] x[/tex] is its distance from the origin and [tex]\beta[/tex] is a constant. A driving force given by [tex]m A \cos{\omega t}[/tex] where [tex]A[/tex] is a constant, is applied to the particle along the x-axis. What value of [tex] \omega[/tex] results in steady-state oscillations about the origin with maximum amplitude? What is the maximum amplitude?

Here's what I've done:

This is the differential equation I've come up with:
[tex]m \ddot{x} + 3 \beta m \dot{x} + \frac{17}{2} \beta^2 m x = m A \cos{\omega t}[/tex]

I'm pretty sure these two steps are the correct ones to make, my textbooks uses this approach:
[tex]A \cos{\omega t} = A e^{i \omega t}[/tex]

[tex]x(t) = A_2 e^{i(\omega t - \theta)[/tex]

This is what I got after plugging in x(t) and solving the derivitives, and dropping the m term:
[tex]-A_2 \omega^2 e^{i(\omega t - \theta)} + A_2 3 i \beta \omega e^{i(\omega t - \theta)} + A_2 e^{i(\omega t - \theta)} = A e^{i \omega t}[/tex]

This is after I eliminated [tex]e^{i \omega t}[/tex] and moved [tex]e^{-i \theta}[/tex]:
[tex]A_2(\frac {17} {2} \beta^2 - \omega^2 + 3 i \beta \omega) = A e^{i \theta}[/tex]

These are the equations I got after applying Euler's identity:
[tex]A_2(\frac {17} {2} \beta^2 - \omega^2) = A \cos{\theta}[/tex]

[tex]A_2 3 \beta \omega = A \sin{\theta}[/tex]

I divided one equation by the other for [tex]\theta[/tex]:
[tex]tan{\theta} = \frac {3 \beta \omega} {\frac {17} {2} \beta^2 - \omega^2}[/tex]

I squared and added both equation's then solved for [tex]A^2_2[/tex]:
[tex]A^2_2 = \frac {A^2} {\omega^4 - 8 \beta^2 \omega^2 + \frac {289} {4} \beta^4}[/tex]

edit: I solved it on my own. Thanks for nothing!
Taking the derivitive of the last equation with respect to [tex]\omega[/tex]:
[tex]A'_2(\omega) = A \omega (\omega^4 - 8 \beta \omega + \frac {289} {4} \beta^4)^{-3/2} (8 \beta^2 - 2 \omega^2)[/tex]

Setting [tex]A'_2 = 0[/tex] and recognizing that [tex]\omega = 0 [/tex] is an arbitrary solution [tex] 8 \beta^2 - 2 \omega^2 = 0 [/tex].
Therefore [tex]\omega_r = 2 \beta[/tex].

For the second part of the problem I plug [tex]\omega_r[/tex] into [tex]A_2(\omega)[/tex]:
[tex]A_2(\omega_r) = A(\frac {225} {4} \beta^4)^{\frac {-1} {2}}[/tex]

The final result!:
[tex]A_2(\omega_r) = \frac {2 A} {15 \beta^2}[/tex]
 
Last edited:

bgc

11
0
drop the m in the RHS. Why did you use it? Also an atrctive force must be negative if you have it on the LHS.

the question implies it's an oscillator (under damped) this will only be true if beta is < 0.

Furthermore, the m's in the forces ar curious.

Since you solved it - I would necessarily check it to find why my suggestions appear invalid.
 

fluidistic

Gold Member
3,625
96
The OP hasn't logged in since more than 4 years, so do not expect a reply from his part.
 

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