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Finding the amplitude an oscillator: Driven harmonic oscillator problem 
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#1
Oct1808, 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]^{2}h(ct) where K and [tex]\Omega[/tex] are positve constants. Suppose that the profile of the road surface is given by h(x) = h_{0}cos(px/c), where h_{0} 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 http://courses.ncsu.edu/py411/lec/001/ 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]=yh(ct) [tex]\varsigma[/tex](single dot)=y(single dot)h'c(ct) y=ce^{ipt} y(single dot)=cipe^{ipt} y(double dot)=c^2e^{ipt} since h(x)=h(ct) and h(x) = h_{0}cos(px/c),then h(x)= h_{0}cos(px/c)= h_{0}cos(pt) h(x)= h_{0}cos(pt) h'(x)=p h_{0}sin(pt) could I say h(x)= h_{0}cos(pt)=h0e^ipt? then h(x)=h0e^ipt h'(x)=iph0e^ipt therefore, [tex]\varsigma[/tex]=yh(ct) becomes [tex]\varsigma[/tex]=yh(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]^{2}h(ct) I find c to be : c=h0(2kp^2+[tex]\Omega[/tex]^{2})/([tex]\Omega[/tex]^{2}+2ki2kh0p) 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} 


#2
Oct1908, 05:39 PM

P: 569

anybody have any trouble reading my solution?



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