# Finding the amplitude an oscillator: Driven harmonic oscillator problem

by Benzoate
Tags: amplitude, driven, harmonic, oscillator
 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)+$$\Omega$$2= 2Kch'(ct) + $$\Omega$$2h(ct) where K and $$\Omega$$ 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 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$$\varsigma$$'+$$\Omega$$2$$\varsigma$$=0 y=h(ct)+$$\varsigma$$==> $$\varsigma$$=y-h(ct) $$\varsigma$$(single dot)=y(single dot)-h'c(ct) y=ceipt 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? then h(x)=h0e^ipt h'(x)=iph0e^ipt therefore, $$\varsigma$$=y-h(ct) becomes $$\varsigma$$=y-h(pt)=> $$\varsigma$$(single dot)=y(single dot)-h'p(pt) plugging all of my variables into the equation y(double dot) + 2Ky(single dot)+$$\Omega$$2= 2Kch'(ct) + $$\Omega$$2h(ct) I find c to be : c=h0(2kp^2+$$\Omega$$2)/($$\Omega$$2+2ki-2kh0p) 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= (($$\Omega$$4+4K^2p^2)/(($$\Omega$$2-p^2)2+4K^2p^2))1/2