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Finding the amplitude an oscillator: Driven harmonic oscillator problem

by Benzoate
Tags: amplitude, driven, harmonic, oscillator
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Oct18-08, 11:56 AM
P: 569
1. The problem statement, all variables and given/known data

A car is moving along a hill at constant speed on an undulating road with profile h(x) where h'(x) is small. The car is represented by a chassis which keeps contact with the road , connected to an upper mass m by a spring and a damper. At time t, the upper mas has displacement y(t) satisfies a differential equation of the form

y(double dot) + 2Ky(single dot)+[tex]\Omega[/tex]2= 2Kch'(ct) + [tex]\Omega[/tex]2h(ct)

where K and [tex]\Omega[/tex] are positve constants.

Suppose that the profile of the road surface is given by h(x) = h0cos(px/c), where h0 and p are positive constants. Find the amplitude a of the driven oscillations of the upper mass.

I will post the website that contains my professor's hint to this problem and since the hint is in pdf form, I am unable to paste it

Go to homework tab
Then go to assignment 7
then go to 5.11 once you've clicked on assignment 7

2. Relevant equations

3. The attempt at a solution

y(double dot)+2K[tex]\varsigma[/tex]'+[tex]\Omega[/tex]2[tex]\varsigma[/tex]=0

y=h(ct)+[tex]\varsigma[/tex]==> [tex]\varsigma[/tex]=y-h(ct)

[tex]\varsigma[/tex](single dot)=y(single dot)-h'c(ct)
y(single dot)=cipeipt
y(double dot)=-c^2eipt
since h(x)=h(ct) and h(x) = h0cos(px/c),then h(x)= h0cos(px/c)= h0cos(pt)

h(x)= h0cos(pt)
h'(x)=-p h0sin(pt)

could I say h(x)= h0cos(pt)=h0e^ipt?


therefore, [tex]\varsigma[/tex]=y-h(ct) becomes [tex]\varsigma[/tex]=y-h(pt)=> [tex]\varsigma[/tex](single dot)=y(single dot)-h'p(pt)

plugging all of my variables into the equation y(double dot) + 2Ky(single dot)+[tex]\Omega[/tex]2= 2Kch'(ct) + [tex]\Omega[/tex]2h(ct)

I find c to be :


I do realize in order to get the amplitude I have to calculate the magnitude of c: I think I calculated my magnitude incorrectly :

According to my textbook , here is the actually amplitude

a= (([tex]\Omega[/tex]4+4K^2p^2)/(([tex]\Omega[/tex]2-p^2)2+4K^2p^2))1/2
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Oct19-08, 05:39 PM
P: 569
anybody have any trouble reading my solution?

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