jsMath

Forced vibrations

       
This example illustrates the relationship between the solution
of a differential equation modeling forced vibration and the forcing function. Here we consider the
initial value problem
y+0.125y+y= 3cos(ωt),y(0)=0,y(0)=2.
In blue is the solution to the IVP, while in red is the forcing function
F(t)=3cos(ωt).
Here you can see how
different values of ω affect the amplitude of the solution.
Here are plots for various values of ω:
ω=0.3 ω=1 ω=3 Here is an animation showing the solution for many values of ω:
       
This example illustrates periodic variation of amplitude which occurs when you have a forced, 
undamped vibration. Suppose a vibration is modelled by
y+y=cos(ωt),y(0)=0,y(0)=0.
Then the solution will be
y=(21ω2sin(2(1ω)t))sin(2(1+ω)t). 
If ω is close to 1, then 1ω  is small, and hence the factor
sin(2(1+ω)t) 
is oscillating much faster than
21ω2sin(2(1ω)t). 
So, we think of | |  21ω2sin(2(1ω)t)| |    (shown in red) as periodically varying the amplitude of the sine curve.
Here are plots for various values of ω:
ω=0.7 ω=0.9 ω=0.95 Here's an animation showing the solution for many values of ω: