Mathematical Tools for Physics

4—Differential Equations 106

With a little thought (i.e. don’t plunge blindly ahead) you can solve these easily.

A 1 =A 2 =A 3 =A 4 =

x 0

4

x 1 (t) =

x 0

4

[

eiω^1 t+e−iω^1 t+eiω^2 t+e−iω^2 t

]

=

x 0

2

[

cosω 1 t+ cosω 2 t

]

x 2 (t) =

x 0

4

[

eiω^1 t+e−iω^1 t−eiω^2 t−e−iω^2 t

]

=

x 0

2

[

cosω 1 t−cosω 2 t

]

From the results of problem3.34, you can rewrite these as

x 1 (t) =x 0 cos

(

ω 2 +ω 1

2

t

)

cos

(

ω 2 −ω 1

2

t

)

x 2 (t) =x 0 sin

(

ω 2 +ω 1

2

t

)

sin

(

ω 2 −ω 1

2

t

)

(35)

As usual you have to draw some graphs to understand what these imply. If the center springk 3 is a
lot weaker than the outer ones, then Eq. ( 32 ) implies that the two frequencies are close to each other and
so|ω 1 −ω 2 | ω 1 +ω 2. Examine Eq. ( 35 ) and you see that one of the two oscillating factors oscillate at
a much higher frequency than the other. To sketch the graph of[ x 2 for example you should draw one factor
sin

(

(ω 2 +ω 1 )t/ 2

)]

and the other factor

[

sin

(

(ω 2 −ω 1 )t/ 2

)]

and graphically multiply them.

x 2

The massm 2 starts without motion and its oscillations gradually build up. Later they die down and build
up again (though with reversed phase). Look at the other mass, governed by the equation forx 1 (t)and you see
that the low frequency oscillation from the(ω 2 −ω 1 )/ 2 part is big where the one forx 2 is small and vice versa.
The oscillation energy moves back and forth from one mass to the other.