MTH 264
DELTA COLLEGE
SECTION 4.3 16
For the following second-order equation
(a) determine the frequency of the beats,
(b) determine the frequency of the rapid oscillations, and
(c) use the information from parts (a) and (b) to give a rough sketch of a typical solution.
y 00 + 11y = 2 cos 3t
Solution:
√
This is an undamped harmonic oscillator, with a natural frequency ω0 = 11. Since the frequency of the
driving function is ω = 3, does not match ω0 , there is no resonance in this problem.
A particular solution of this equation will involve the product of two sine waves of differing frequencies of the
form:
ω − ω0
ω + ω0
t sin
t
sin
2
2
The frequency of the rapid oscillations is the average of the natural and driving frequencies and the frequency
of the beats is the difference (in absolute value) between the two frequencies and their average.
√
ω − ω0
3 − 11
beat frequency
=
≈ 0.1583 cycles per 2π seconds ≈ 0.0252 Hz
2
2
rapid oscillation frequency
√
ω + ω0
3 + 11
=
≈ 3.1583 cycles per 2π seconds ≈ 0.5027 Hz
2
2
We are interpreting frequency as “cycles per 2π seconds”, which is commonly called angular frequency. Divide
both of these numbers by 2π if you want to calculate the number of cycles per second (Hz).
The ratio of the rapid oscillation frequency to the beat frequency will give you a rough idea of how many
rapid oscillations occur within one beat oscillation.
√
ω + ω0
2+ 6
√ ≈ 20.0
=
ω − ω0
2− 6
Below is a rough picture of two beats. In the graph, it may be easier for you to relate to the period of the
two oscillations
beat period
4π
4π
√ ≈ 39.7 seconds
=
ω − ω0
3 − 11
rapid oscillation period
4π
4π
√ ≈ 2.0 seconds
=
ω + ω0
3 + 11