Forced Oscillation
Problem: Suppose a water tower in an earthquake acts as a mass-spring system. Assume that the container on top
is full and the water does not move around. The container then acts as a mass and the support acts as the spring,
where the induced vibrations are horizontal. Suppose that the container with water has a mass of m = 10, 000
kg. It takes a force of 1000 N to displace the container 1 meter. For simplicity assume no friction. When the
earthquake hits the water tower is at rest (it is not moving). Suppose that an earthquake induces an external force
F (t) = mAω 2 cos(ωt).
(a.) What is the natural frequency of the water tower?
(b.) If ω is not the natural frequency, find a formula for the maximal amplitude of the resulting oscillations
of the water container (the maximal deviation from the rest position). The motion will be a high frequency wave
modulated by a low frequency wave, so simply find the constant in front of the sines.
(c.) Suppose A = 1 and an earthquake with frequency 0.5 cycles per second comes. What is the amplitude of the
oscillations? Suppose that if the water tower moves more than 1.5 meter from the rest position, the tower collapses.
Will the tower collapse?
Solution: (a.) Note that the fact that it takes 1000 N to displace the container 1 meter implies that the stiffness of
the “spring” (which in this case is actually the water tower support) is k = 1000 N/m. Therefore, the mass-spring
system is given by:
10000x00 + 1000x = mAω cos(ωt)
The natural frequency of the water tower is the frequency with which it would oscillate if set in motion in the
absence of any external force. Thus, we need to solve the homogeneous equation:
10000x00 + 1000x = 0.
1
The roots of the characteristic equation are ±i √ and so the solution to the homogenous equation is:
10
1
1
C1 cos √ t + C2 sin √ t
10
10
r
1
k
from which we see that the natural frequency is ω0 = √ rad/sec . Note that ω0 =
. This is formula holds in
m
10
generally for mass-spring systems in which friction is ignored.
(b.) Suppose ω (the frequency of the external force induced by the eaarthquake) is not equal to ω0 . By using the
method of undetermined coefficients, one can solve
mx00 + kx = F0 cos(ωt)
to obtain the general solution:
x = C1 cos(ω0 t) + C2 sin(ω0 t) +
r
where ω0 =
F0
cos(ωt).
− ω2 )
m(ω02
k
. We can use the general trigonometric identity:
m
p
A cos(ω0 t) + B sin(ω0 t) = A2 + B 2 cos(ω0 t − γ),
γ = arctan
to rewrite the general solution as:
x=
q
C12 + C22 cos(ω0 t − γ) +
F0
cos(ωt).
m(ω02 − ω 2 )
B
A
Now, if we assume that our system is initially at rest so that x(0) = 0 and x0 (0) = 0, then we can solve for C1 and
C2 in the homogenous solution to see that
C1 = −
Then the solution becomes:
x=
F0
,
m(ω02 − ω 2 )
C2 = 0.
F0
(cos(ωt) − cos(ω0 t))
− ω2 )
m(ω02
We can again use an obscure trigonometric identity for the difference of two cosines:
A−B
A+B
cos(B) − cos(A) = 2 sin
sin
2
2
to arrive at:
x=
(ω0 − ω)t
2F0
(ω0 + ω)t
sin
sin
m(ω02 − ω 2 )
2
2
To be clear about our assumptions, the above is the solution to the mass-spring system:
mx00 + kx = F0 cos(ωt),
ω 6= ω0 ,
x(0) = 0, x0 (0) = 0
q
1
In our water tower problem, we have F0 = mAω 2 , ω0 =
10 . Note that our problem is only asking for the
maximum amplitude of oscillation which from the general solution above is given by:
2Aω 2 20Aω 2 2F0
=
=
maximum amplitude = 1
m(ω02 − ω 2 ) 10
− ω 2 1 − 10ω 2 (c.) Now we assume that A = 1 and that ω = 0.5 cycles per second. Note that we need to convert ω so that it
has units of radians per second, so:
cycles
rad
rad
ω = 0.5
· 2π
=π
sec
cycle
sec
Then the max amplitude of oscillations is:
20π 2 ≈ 2.02 meters .
max amp. = 1 − 10π 2 Evidently, the tower will collapse if such an earthquake occurs.
Some comments and pictures
1
Suppose that ω starts getting closer to the value of ω0 = √
≈ 0.3162. What happens to the amplitude of
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oscillation? From our formula for maximum amplitude above we see that the denominator gets closer to 0 and so
the maximum amplitude blows up.
What about the displacement x of the water tower itself? I have plotted the graph of x for values of ω =
0.5, 0.4, 0.35. These plots are Figures 1-3 below.
Remember:
(ω0 − ω)t
(ω0 + ω)t
2F0
sin
sin
.
x=
2
2
m(ω0 − ω )
2
2
So what we see is a sine wave given by the right-hand sine term amplified and attenuated by the longer-period
smaller-period sine wave inside the brackets. This sine wave in the brackets acts as envelope curves. In the plots
given below, I have drawn these envelopes in red.
Figure 1: ω = 0.5, the maximum amplitude is 3.33
Figure 2: ω = 0.4, the maximum amplitude is 5.33
Figure 3: ω = 0.35, the maximum amplitude is 10.88
Figure 4: ω = ω0
Q: What about if ω = ω0 ? A: Resonance
1
The natural frequency of the water tower is ω0 = √ . Suppose that we apply an external force which is periodic
10
with the exact same frequency ω = ω0 . This will give rise to a physical phenomenon called resonance.
We can see right off the bat that the above derivation won’t work because the guess made for xp , namely A cos(ω0 t)
is actually a solution to the homogeneous equation. Therefore we modify our guess to:
xp = At cos(ω0 t)
Working through the messy details, we arrive at the general solution:
x = C1 cos(ω0 t) + C2 sin(ω0 t) +
F0
t sin(ω0 t)
2mω0
If we assume the system is initially at rest so that x(0) = 0 and x0 (0) = 0, then we can see C1 = 0, and since:
x0 (t) = C2 ω0 cos(ω0 t) +
F0
F0
sin(ω0 t) +
t cos(ω0 t)
2mω0
2m
so C2 is also 0.
Thus, the solution to the initial value problem:
x00 + ω02 x = F0 cos(ω0 t),
x(0) = 0,
x0 (0) = 0
is
x=
F0
t sin(ω0 t).
2mω0
Thus we have oscillation, but we see that the amplitude of oscillation increases linearly with time. Figure 4 is the
plot of resonance in the case of our water tower subjected to the earthquake (now of frequency ω0 ). Note that the
envelope curves in red are now lines. Can you kind of see how letting ω limit towards ω0 produces this phenomenon?
In any case, as we can see, resonance is very bad for water towers.