60 1 Mathematical Physics
The equation of motion for mass 2 is
mx ̈ 2 +k(2x 2 −x 1 )=0(2)
The two Eqs. (1) and (2) are coupled equations.
Letx 1 =A 1 sinωt (3)
x 2 =A 2 sinωt (4)
x ̈ 1 =−ω^2 A 1 sinωt=−ω^2 x 1 (5)
x ̈ 2 =−ω^2 A 2 sinωt=−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 ofx 1 andx 2 must vanish.
∣
∣
∣
∣
∣2 k−mω^2 −k
−k 2 k−mω^2∣
∣
∣
∣
∣
=0(9)
(2k−mω^2 )^2 −k^2 = 0
or (mω^2 −k)(mω^2 − 3 k)= 0
The solutions areω 1 =√
k
m(10)
ω 2 =√
3 k
m(11)
(b) The two solutions to the problem are
x 1 =A 1 sinω 1 t;x 2 =A 2 sinω 1 t (12)
x 1 =B 1 sinω 2 t;x 2 =B 2 sinω 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), yieldsA 2 =
A 1. Substitution of (11) and (13) in (7), givesB 2 =−B 1.
Dropping off the subscripts on A′s and B′s the solutions can be
written as
x 1 =Asinω 1 t=x 2 (14)
x 1 =Bsinω 2 t=−x 2 (15)