60
1 Mathematical Physics
The equation of motion for mass 2 is
m ¨
x 2 + k(2x 2 − x 1 ) = 0
( 2 )
The two Eqs. (1) and (2) are coupled equations.
Let x 1 = A 1 sin ωt
(3)
x 2 = A 2 sin ωt
(4)
¨
x 1 = −ω
2 A 1 sin ωt = −ω
2 x 1
(5)
¨
x 2 = −ω
2 A 2 sin ωt = −ω
2 x 2
(6)
Inserting (5) and (6) in (1) and (2)
−mω
2 x 1 + k(2x 1 − x 2 ) = 0
− mω
2 x 2 + k(2x 2 − x 1 ) = 0
Rearranging
(2k − mω
2 )x 1 − kx 2 = 0
( 7 )
−kx 1 + (2k − mω
2 )x 2 = 0
( 8 )
In order that the above equations may have a non-trivial solution, the
determinant formed from the coefficients of x 1 and x 2 must vanish.
2k − mω
2
−k
−k
2k − mω
2
= 0
( 9 )
(2k − mω
2 )
2
− k
2
= 0
or (mω
2
− k)(mω
2
− 3k) = 0
The solutions are
ω 1 =
k
m
(10)
ω 2 =
3k
m
(11)
(b) The two solutions to the problem are
x 1 = A 1 sin ω 1 t; x 2 = A 2 sin ω 1 t
(12)
x 1 = B 1 sin ω 2 t; x 2 = B 2 sin ω 2 t
(13)
In (12) and (13) the amplitudes are not all independent as we can verify
with the use of (7) and (8). Substituting (10) and (12) in (7), yields A 2 =
A 1 . Substitution of (11) and (13) in (7), gives B 2 = −B 1 .
Dropping off the subscripts on A
s and B
s the solutions can be
written as
x 1 = A sin ω 1 t = x 2
(14)
x 1 = B sin ω 2 t = −x 2
(15)