Resonance
In every day life resonance can be a fairly common phenomenon although you may
not realise it. Resonance occurs when a system is forced at its natural frequency,
leading to a build-up of the amplitude of oscillation and energy. The effect is familiar
to most as the high pitched squeal over a PA system caused by microphone feedback.
Mathematically, resonance can be seen as a non-homogeneous second order
differential equation whose particular solution is of the same form as the
complimentary function. To see how this happens, let us look at a simple example:
d2y
 4 y  3 cos2t .
dt
To solve this problem let’s start by finding the complimentary function. The auxiliary
equation is
m2  4  0,
which gives us a pure complex root m  2i . Therefore we know the complimentary
function is:
yc  A cos2t   B sin 2t .
Now when we come to the particular solution we have a problem. As the RHS
contains cos2t  , our guess at a particular solution would be:
y p  A cos2t   B sin 2t ,
However as this is the same as our complimentary solution we will find that
substituting this into the differential equation well result in a zero answer. To illustrate
this point let’s try the substitution.
First we work out the first and second differentials of y p , giving:
y p '  2 A sin 2t   2 B cos2t ,
y p ' '  4 A cos2t   4 B sin 2t .
Substituting this into the LHS of the question gives:
 4 A cos2t   4B sin 2t   4 A cos2t   B sin 2t   3cos2t .
As expected, all the terms on the LHS will cancel down to zero. Thus our guess is
never going to lead to the complete answer.
For an idea of how to solve this problem, let’s look at how we think resonance should
behave. If a system is in resonance we expect the amplitude of the vibration to
increase as the forcing stores energy in the system. The complimentary function has a
constant amplitude so it seems reasonable that the particular solution may have an
amplitude which increases with time. The simplest way to do this is multiply our
original guess by t , i.e. we guess a new particular solution:
y p  At cos2t   Bt sin 2t .
This time when we differentiate we must use the product rule. Calculating the first
differential, we get:
y p '  A cos2t   2 At sin 2t   B sin 2t   2 Bt cos2t .
The second differential gets a little lengthy as we must apply the product rule two
more times:
y p ' '  2 A sin 2t   2 A sin 2t   4 At cos2t   2 B cos2t   2 B cos2t   4 Bt sin 2t ,
collecting like terms give us a more manageable expression:
y p ' '  4 A sin 2t   4 B cos2t   4 At cos2t   4 Bt sin 2t .
This time when we substitute the expression into the original problem something
different happens. The terms containing t cos2t  and t sin 2t  cancel out, as shown
below:
 4 Asin 2t   4B cos2t   4 At cos2t   4Bt sin 2t   4 At cos2t   Bt sin 2t   3cos2t .
Performing the cancelations gives:
 4 A sin 2t   4B cos2t   3 cos2t .
Now we are able to equate coefficients:
 4 A  0,
4B  3.
Hence we find our particular solution is:
yp 
3t
sin 2t ,
4
which, as we expected, has an increasing amplitude in time. The full general solution
is:
y  A cos2t   B sin 2t  
3t
sin 2t .
4
We could also be given initial conditions allowing us to solve for A and B .
In general it is always best to solve for the complimentary solution first; that way you
can spot if you will have a problem with resonance. If you attempt to solve for the
particular solution first and find that your guess leads to a nonsense answer such as
0  f t  , then you’ve probably got a resonance question. To deal with the problem
simply multiply your solution by the dependant variable, i.e. x or t . We consolidate
this idea with some examples:
Example 1:
Q) We would like to solve the non-homogeneous equation:
d2y
 9 y  3 cos3t   2 sin 3t 
dt 2
A)
We start by solving the homogeneous equation to find the complimentary solution:
d 2 yc
 9 yc  0.
dt 2
Form the auxiliary equation:
m2  9  0,
giving us pure complex roots m  3i . Therefore the complimentary solution is:
yc  A cos3t   B sin 3t .
We can now tell we are going to have a resonance problem as the complimentary
solution has the same form as the forcing term on the RHS of the original problem.
Therefore we need to multiply by t to form our guess at the particular solution:
y p  At cos3t   Bt sin 3t .
We will leave the detail of how you complete the first and second differentials as an
exercise, but record the results here for reference:
y p '  A cos3t   B sin 3t   3Bt cos3t   3 At sin 3t ,
y p ' '  6 B cos3t   6 A sin 3t   9 At cos3t   9 Bt sin 3t .
Substituting this particular solution into the equation gives:
6B cos3t   6 A sin 3t   3 cos3t   2 sin 3t ,
where we have already performed the obvious cancellation of terms. It’s fairly simple
to see that we require A  1/ 3 and B  1 / 2 , giving us a particular solution:
yp 
t
t
cos3t   sin3t ,
3
2
and a full general solution:
t
t
y  y c  y p  A cos3t   B sin 3t   cos3t   sin 3t .
3
2
Example 2
Q) In this example we will show that it is possible to have an exponential resonance.
Let’s try and solve:
d 2 y dy
  2 y  5et
2
dt
dt
A)
As always, we will start by looking at the homogeneous equation:
d 2 y c dy c

 2 y c  0.
dt 2
dt
This leads to the auxiliary equation:
m2  m  2  0,
which gives us roots m1  1 and m2  2 . The complimentary function is then:
yc  Aet  Be 2t .
Therefore if we use Ae t as our guess at the particular solution we will find this is
reduced to zero by the differentials. Therefore we multiply this by t and use:
y p  Atet
as our guess for the particular integral. Again you will need to take some care in
performing differentiation as you will need to apply the product rule.
Here are the first and second differentials:
y p '  Aet  Atet ,
y p ' '  2 Aet  Atet .
Putting these back into the original equation, we find:
 2 Aet  Atet  Aet  Ate t   2 Ate t  5e t .
After expanding out and cancelling we get:
 3 Ae  t  5e  t .
As expected, all the terms containing te t have cancelled. We can now see that
A  5 / 3 , giving a particular solution of:
yp  
5t  t
e ,
3
and full solution:
y  Aet  Be 2t 
5t t
e .
3
Example 3
Q)
Find the general solution of:
d2y
dy
 2  2 y  et 3 cost   sin t 
2
dt
dt
A)
In a procedure that should now be very familiar we take the homogeneous equation:
d 2 yc
dy
 2 c  2 y c  0,
2
dt
dt
and form the auxiliary equation:
m2  2m  2  0,
and find this give us complex roots 1  i . We find the complimentary solution to be:
yc  e  t  A cost   B sin t .
As we expected for a resonance question the complimentary solution is of the same
form as the forcing term. Therefore we multiply by t and use this as the guess for our
particular solution:
y p  tet  A cost   B sin t .
With care the differentials can be found. Applying the product rule once gives:
y p '  et  A cost   B sin t   tet  A cost   B sin t   tet  Asin t   B cost .
Differentiating a second time gives:
y p ' '  2et  A cost   B sin t   2et  A sin t   B cost   2tet  A sin t   B cost .
For the second differential we have collected up like terms and performed a
cancellation. Putting this all together in the original equation, we find:
2et  A sin t   B cost   et 3 cost   sin t .
Again all the terms have cancelled with each other apart from two and we can read off
a solution of A  1/ 2 and B  3 / 2 , giving us our particular solution:
1
3

y p  te t  sin t   cost ,
2
2

and full general solution:
1
3

y  e  t  A cost   B sin t   te t  sin t   cost .
2
2

Example 4
Q)
Sometime we’ll have a mixture of forcing terms, some that are in resonance and some
that are not. For instance, let’s look at:
d2y
 25 y  2 sin 5t   cost .
dt 2
A)
First we solve the homogeneous equation:
d 2 yc
 25 y c  0,
dt 2
with the auxiliary:
m 2  25  0.
It should come as little surprise that we have pure complex roots m  5i and a
complimentary solution of:
yc  A cos5t   B sin 5t .
Now looking back at the question we can see that the first term of the RHS is in
resonance with the system, however the second term is not. Therefore we use a copy
of the complimentary solution multiplied by t added to a general mix of cost  and
sin t  as our guess for the particular solution:
y p  At cos5t   Bt sin 5t   C cost   D sin t .
Working out the first and second differentials:
y p '  A cos5t   B sin 5t   5 At sin 5t   5Bt cos5t   C sin t   D cost ,
y p ' '  10 A sin 5t   10 B cos5t   25 At cos5t   25Bt sin 5t   C cost   D sin t .
On substituting into the equation we expect all the t cos5t  and t sin 5t  terms to
cancel:
10 Asin 5t   10B cos5t   24C cost   24D sin t   2 sin 5t   cost .
We can see that both B  0 and D  0 , therefore we don’t need these terms in the
particular solution. If we had been clever and thought ahead we may have been able to
foresee this and save ourselves some work. This sort of insight should hopefully come
with practice; if in doubt include all the terms. From above we can also see A  1 / 5
and C  1 / 24 giving us:
yp 
1
t
cost   cos5t ,
24
5
and full solution:
y  A cos5t   B sin 5t  
1
t
cost   cos5t .
24
5