A Function for Damped Simple Harmonic Motion

Date: 10/17/98 at 23:36:08
From: William Liao
Subject: Periodic and exponential function

Dr. Maths,

I have problem understanding the following problem in my textbook::

The end of a plastic ruler is vibrating in damped harmonic motion.  
When it starts vibrating, the amplitude is 0.9cm, but 2 seconds later 
has decreased to 0.4cm. In this time it has moved backwards and 
forwards 100 times. Find an expression for the position of the end of 
the ruler, taking the centre of vibration as 0.

In the solution given for this sxample, I couldn't understand the part 
where it said: "In 2 seconds, 100 vibrations are completed, so the 
frequency is 50/second and the period is thus 0.02s, giving b = 50 x 
2pi." (y = Ae(at) cos bt) Also I don't quite see how it said: "After 2 
seconds the amplitude is only 0.4 cm, so 0.4 = 0.9e(2a)."

Please try to explain it to me. Thank you.

Sincerely,
Will

Date: 10/18/98 at 17:54:55
From: Doctor Anthony
Subject: Re: Periodic and exponential function

The equation for damped SHM (simple harmonic motion) is:

   x = Ae^(-kt)cos(wt)   where A, k, w are constants to be found.

When t = 0, x = 0.9 so 0.9 = A, and we have:

   x = 0.9e^(-kt)cos(wt)

In 2 seconds it completes 100 oscillations and so does 50 cycles per 
second:

   Period = 2pi/w  = 1/50   and   w = 100pi
   x = 0.9e^(-kt)cos(100.pi.t)

When t = 2, x = 0.4, so:

   0.4 = 0.9e^(-2k)cos(200pi)
   4/9 = e^(-2k)
   2/3 = e^(-k)
   -k = ln(2/3) = -0.405  
    k = 0.405

Thus we get:

   x = 0.9e^(-0.405t)cos(100pi t)

This is the equation of the damped simple harmonic motion.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/