# Driven Damped Harmonic Oscillator Problem

Walczyk
A mass m moves along the x-axis subject to an attractive force given by $$\frac {17} {2} \beta^2 m x$$ and a retarding force given by $$3 \beta m \dot{x}$$, where $$x$$ is its distance from the origin and $$\beta$$ is a constant. A driving force given by $$m A \cos{\omega t}$$ where $$A$$ is a constant, is applied to the particle along the x-axis. What value of $$\omega$$ 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:
$$m \ddot{x} + 3 \beta m \dot{x} + \frac{17}{2} \beta^2 m x = m A \cos{\omega t}$$

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

$$x(t) = A_2 e^{i(\omega t - \theta)$$

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

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

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

$$A_2 3 \beta \omega = A \sin{\theta}$$

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

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

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

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

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

The final result!:
$$A_2(\omega_r) = \frac {2 A} {15 \beta^2}$$

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