Exponentially driven harmonic oscillator

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Homework Help Overview

The problem involves an un-damped harmonic oscillator with a natural frequency subjected to an exponentially driven force. The objective is to find the equation of motion given specific initial conditions.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of the differential equation and the method for finding the particular solution. There is an exploration of the relationship between the coefficients and the terms in the equation.

Discussion Status

Some participants have identified a potential typo in the differential equation and are verifying the approach to finding the particular solution. There is a recognition of the need to apply initial conditions to determine unknown constants.

Contextual Notes

Initial conditions are specified as both position and velocity being zero at time t=0. There is an emphasis on ensuring dimensional consistency in the differential equation.

vbrasic
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Homework Statement


An un-damped harmonic oscillator natural frequency ##\omega_0## is subjected to a driving force, $$F(t)=ame^{-bt}.$$ At time, ##t=0##, ##x=\dot{x}=0##. Find the equation of motion.

Homework Equations


##F=m\ddot{x}##

The Attempt at a Solution


We have $$m\ddot{x}+kx=F(t)=ame^{-bt}.$$ Dividing through by ##m## we have $$\ddot{x}+\omega_0^2x=ae^{-bt}.$$ From a course in differential equations, I know that the solution to this is the sum of the homogenous solution with the particular solution. The homogenous solution is of the form $$A\cos{\omega_0t}+B\sin{\omega_0t}.$$ I am having a bit of trouble finding the particular solution. Because the solution is an exponential, I assume a solution of the form ##x(t)=ce^{-bt}.## Differentiating twice, we have $$\ddot{x}=b^2ce^{-bt}.$$ Plugging into the expression gives, $$b^2ce^{-bt}+\omega_0^2ce^{-bt}=ae^{-bt}.$$ From this we have that $$a=b^2c+\omega_0^2c.$$ I'm not sure if I'm on the right track here. Any help would be appreciated.
 
Last edited:
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First of all, you have a typo in your differential equation. I suggest you check the dimensions of each term to find it.

vbrasic said:
Plugging into the expression gives, $$b^2ce^{-bt}+\omega_0ce^{-bt}=ae^{-bt}.$$ From this we have that $$a=b^2c+\omega_0c.$$ I'm not sure if I'm on the right track here. Any help would be appreciated.

What is your unknown in finding the particular solution? Can you find it from your equation? Does what you get solve the differential equation?
 
Orodruin said:
First of all, you have a typo in your differential equation. I suggest you check the dimensions of each term to find it.

Should have been ##\omega_0^2.## Also, I'm not solving for ##a##. Rather I should be solving for the ##c## of the particular solution. So from the $$(b^2+\omega_0^2)c=a,$$ we get $$c=\frac{a}{b^2+\omega_0^2}.$$ We verify this is indeed the particular solution. So the most general solution is, $$A\cos{\omega_0t}+B\sin{\omega_0t}+\frac{a}{b^2+\omega_0^2}e^{-bt}.$$ From here we would apply the initial conditions?
 
Indeed, applying the initial conditions will fix your unknown constants ##A## and ##B##.
 

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