3.9  Forced Mechanical Vibrations

Sometimes it is useful to model a damped mechanical vibration with an inhomogeneous differential equation
                mu''  +  γu'  +  ku  =  g
where the function g accounts for some extra force applied to the vibrating object .   A good example is
      that of a car (the object) moving on top of its shock absorbers (the damped spring) on a bumpy road \(\((
        the\ bumps\ compress\ the\ shock\ absorbers\ an\ extra
amount\ and\ 
          therefore\ exert\ an\ extra\ force\ on\ the\ car)\) . \)

\(
Here\ we\ consider\ two\ \(examples : \ one\ with\ no\ damping\)\ 
    \)\(\((γ\  = \ 0)\) and\ another\ with \((under)\) damping\ 
      \((γ\  = \ 1)\) . \ \ \ 
The\ first\ example\ is\ designed\ to\ 
      illustrate\ \)phenomena such as beat and resonance . The example revisits resonance but also  highlights the role of \(the\ particular\ solution\ 
      \((versus\ the\ solution\ of\ the
homogeneous\ equation)\)\ in\ the\ 
      general\ solution\ of\ an\ \)inhomogeneous linear equation .

In both examples,  we will assume that m = 1 and that\(the\ system\ is\ subject\ to\ a\ periodic\ 
      \((i . e . \ of\ the\ form
`cos[ωt]\ ')\)\ external\ 
      force . \ \ We\ will\ also\ set\ the\ \)initial conditions u[0]  =  3,  u'[0]  =  0.

1.  No damping .

We will work with the equation:
		
     u''  +  25 u  ==  a Cos[ωt]		
\(where\ a\ and\ ω\ will\ take\ on\ values\ specified\ 
        below . \ \ 
Let\ us\ first\ compute\ and\ plot\ the\ solution\ when
        \) the right side is identically zero .

Clear[u]

\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ 0, 	u[0]\  == \ 3, \ 
          \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u1  =  u[t] /. First[soln]

Plot[u1,  {t, 0,  20},  PlotPoints  ->  1000]

Now we introduce a (relatively) slowly varying external force, again solving the differential equation and plotting the solution .

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ \ Cos[t], 	
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u2  =  Expand[u[t] /. First[soln]]

Plot[u2,  {t, 0,  20},  PlotPoints  ->  1000]

...  not much difference .   Let' s try increasing the magnitude of the force :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ \ 5\ Cos[t], 	
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u3  =  Expand[u[t] /. First[soln]]

Plot[u3,  {t, 0,  20},  PlotPoints  ->  1000]

...  still not all that different - just a bit of wavering in the original solution .   Suppose we make the external force vary (relatively)rapidly rather than slowly .   We' ll increasethe magnitude again for the sake of drama :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ \ 500  Cos[25  t], 	
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u4 =  Expand[u[t] /. First[soln]]

Plot[u4,  {t, 0,  20},  PlotPoints  ->  1000]

This introduces a bit of a wiggle into the solution (as you mightexpect),  but still we' re not getting our effort' s worth out of this external force .   The real trick is to make the period of the force nearly equal to the period of the unforced solution .

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ \ 5\ Cos[4.5  t], 	
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u5 =  Expand[u[t] /. First[soln]]

Plot[u5,  {t, 0,  20},  PlotPoints  ->  1000]

That' s a bit more impressive .   Let' s try making the periods even close and plot the solution for a longer time :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ \ 5\ Cos[4.9  t], 	
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u6 =  Expand[u[t] /. First[soln]]

Plot[u6,  {t, 0,  100},  PlotPoints  ->  1000]

This is really suggestive .   Our solution has developed a `beat' that varies much more slowly than either the unforced oscillation or the externalforce .    The mathematical reason for the beat is theformula
                  2 sin ((B - A)/2) sin ((B + A)/2) = cos[A] + cos[B].

			This formula applies directly to the solution u6 in our last example *after* we eliminate theeffects of the unforced solution .   That is,
   
      u6 = u1  -  4.101010 ...  sin[ .05 t]sin[4.95 t].

Note that the second term contains a slowly varying factor (the beat) and a factor which varies at about the same rate as the external force .   Note also that the overall amplitude of the solution has grown from about 3 to over 6.   Before moving on to damped examples, we consider the case where the external force has a period exactly matching that of the unforced solution :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ 25\ u[t]\  == \ \ \ 5  Cos[5  t], 	
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        \)

u7 =  Expand[u[t] /. First[soln]]

Plot[u7,  {t, 0,  100},  PlotPoints  ->  1000]

\(\(This\ is\ called\ the\ resonant\ case . \ \ \(Why?\)
			





    \)\)

2.  An Underdamped System with forcing .

Now we will modify the first equation,  adding adamping term:
		
     u''  +  5 u'  + 25 u  ==  a Cos[ωt]		
where,  as before,  we will experiment with different values of a and ω .    First we compute 
        and plot the solution when the right side is identically zero .

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ \(1\/2\) \(u'\)[t] + 25\ u[t]\  == \ 0, 
          	u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, 

        		\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u[t], \ t]
        
\)

u1  =  u[t] /. First[soln]

Plot[u1,  {t, 0,  20},  PlotPoints  ->  1000,   
      PlotRange  ->  {-3, 3}]

\(
Now\ we\ apply\ a\ periodic\ external\ force\ that\ is\ pretty\)much out of sync with the period of the unforced system :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ \(1\/2\) \(u'\)[t] + 25\ u[t]\  == \ 
            2  Cos[3  t], 	

          			\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
          u[0]\  == \ 3, \ \(u'\)[0]\  == \ 0}, \ u[t], \ t]
\)

u2  =  Simplify[u[t] /. First[soln]]

Plot[u2,  {t, 0,  20},  PlotPoints  ->  1000,   
      PlotRange  ->  {-3, 3}]

It might help to increase the domain a bit here :

Plot[u2,  {t, 0,  40},  PlotPoints  ->  1000,   
      PlotRange  ->  {-3, 3}]

So,  as you might expect,  
        the solution doesn' t dieoff completely .   After the initial vibration settlesdown, there remains vibration caused by the forcing .   Perhaps more surprisingly,  this `longterm' vibration is completely independent of theinitial condition .   That is, if we try a differentinitial condition :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ \(1\/2\) \(u'\)[t] + 25\ u[t]\  == \ 
            2  Cos[3  t], 	

          			\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
          u[0]\  == \ \(-2\), \ \(u'\)[0]\  == \ 1}, \ u[t], \ t]
\)

u3  =  Simplify[u[t] /. First[soln]]

and compare with the solution corresponding tothe first set of initial conditions :

Simplify[u3  -  u2]

we see that the difference goes to zero as t increases .   Mathematically speaking,  this merelyserves to illustrate the principle that any twosolutions of a linear,  inhomogeneous equation differ by a solution of the corresponding\(homogeneous\ \(equation . \)
\)

A final point worth making is that damped systems have\(resonances\ just\ like\ undamped\ systems . \ \ However, 

    the\ eventual\ solution\ of\ a\ resonant\ damped\ system\)will not grow forever :

Clear[u]\(soln\  = \ 
      DSolve[{\ \(\(u'\)'\)[t]\  + \ \(1\/2\) \(u'\)[t] + 25\ u[t]\  == \ 
            2  Cos[\(\(\@399\) t\)\/4], 	

          			\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
          u[0]\  == \ \(-2\), \ \(u'\)[0]\  == \ 1}, \ u[t], \ t]
\)

u4  =  Simplify[u[t] /. First[soln]]

Plot[u4,  {t, 0,  40},  PlotPoints  ->  1000,   
      PlotRange  ->  {-3, 3}]