Expert: Paul Klarreich - 5/12/2009

Question
The suspension in a car acts like damped harmonic oscillator,that is the oscillations in the suspension rapidly die down with time.A model for this includes both exponential and trigonometric function.Suppoe the displacement in a car's suspension is given by this formula.
s(t) = e^(-t/4).Cos(4t)
Show by the direct substitution that the displacement satisfies the differential equation
(6(d^2.s)/dt^2)+ 8ds/dt + 11s = 0

I saw this problem in one of the old exam paper and thought i should learn.Help are much appreciate.Thanks in advance.

Answer
Questioner:   mike
Country:  Australia
Category:  Calculus
Private:  No
 
Subject:  differential equation problem
Question:  The suspension in a car acts like damped harmonic oscillator,that is the oscillations in the suspension rapidly die down with time.A model for this includes both exponential and trigonometric function.Suppoe the displacement in a car's suspension is given by this formula.
s(t) = e^(-t/4).Cos(4t)
Show by the direct substitution that the displacement satisfies the differential equation
(6(d^2.s)/dt^2)+ 8ds/dt + 11s = 0

I saw this problem in one of the old exam paper and thought i should learn.Help are much appreciate.Thanks in advance.
....................................
Hi, Mike,

The problem contains its own solution: "Show by direct substitution"  means do it!


s(t) = e^(-t/4).Cos(4t)

I'll condense a bit:

6s''+ 8s' + 11s = 0

So compute the first derivative (you will need the product rule) and then the second derivative (yes, again), then substitute them, simplify, and it all should come out to zero.

Little trick for putting it together: You will have a factor of e^(-t/4) in all terms.  Delete it at the end.

Abbreviations:  E = e^-t/4,   S = sin(4t),  C = cos(4t)
......................................................
HOWEVER, this does not work.

You must have a miscopying here.  The equation you gave has

s(t) = e(-2t/3) cos (5sqrt(2) t/6)

as its solution.

To solve this differential equation:

6s''+ 8s' + 11s = 0

assume the solution is  e^zt, where z is some (complex) constant.

Then diff & subst:

s' = zs
s'' = z^2 s

and you get the CHARACTERISTIC EQUATION:

6z^2 + 8z + 11 = 0

which you solve, using the Quadratic Formula:

   - 8 +- sqrt(64 - 4(11)(6) )
z = ---------------------------
              12

   - 8 +- sqrt(64 - 264 )
z = ---------------------------
              12

   - 8 +- sqrt(- 200)
z = --------------------
              12

   - 8 +- 10 sqrt(- 2)
z = ------------------
              12

   - 4 +- 5 i sqrt(2)
z = ------------------ = x + iy
           6

The real part, x, goes in the e(xt) factor, and
the imaginary part, y, goes in the cosine factor:

s = e(-2t/3) cos (5sqrt(2) t/6)

So check it over.