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Benzoate

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## Homework Statement

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/ [Broken]

Go to homework tab

Then go to assignment 7

then go to 5.11 once you've clicked on assignment 7

## Homework Equations

## 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=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]=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]

^{2}h(ct)

I find c to be :

c=h0(2kp^2+[tex]\Omega[/tex]

^{2})/([tex]\Omega[/tex]

^{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= (([tex]\Omega[/tex]

^{4}+4K^2p^2)/(([tex]\Omega[/tex]

^{2}-p^2)

^{2}+4K^2p^2))

^{1/2}

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