- #1

Walczyk

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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]

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]

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