Physics
Content
Two Coupled Harmonic Oscillators M. Fatih Yaman
I.
SOLVING TWO COUPLED HARMONIC OSCILLATOR PROBLEM
Starting with te idea of a couple of masses , m1 and m2 , connected together with spring constants k1 , k2 and k3 . Two springs,k1 and k3 , connected to the stationary walls and the third one, k2 , connected between two masses. Equilibrium of the two masses are located in fig 1 at their centers.
k1 + k2 − m1 w2 k2 = (7) 2 k3 + k2 − m2 w k2 After the necessary simplification the equation goes to m1 m2 w4 − w2 (m2 k1 +m2 k2 + m1 k2 + m1 k3 ) + k3 k1 + k3 k2 + k2 k1 = 0
(8) (9)
If the coefficents are subsituted with a,b,c then; m1 .m2 = a m2 .k1 + m2 .k2 + m1 .k2 + m1 .k2 + m1 .k3 = b FIG. 1.
k3 .k1 + k3 .k2 + k2 .k1 = c The idea is these masses are going to stretch x1 and x2 from their equilibrium position. The forces on masses are going to be related to x1 and x2 . The force on the m1 , F1 , is equal to
Having the equations been solved s w1 =
k2 (x2 − x1 ) − k1 x1 = m1 x ¨1
(1)
s w1 = (2)
The resulting motion has an oscillatory behaviour, so the equations of x1 and x2 are
√
b2 − 4ac 2a
(10)
b2 − 4ac 2a
(11)
and
and on the m2, F2, is equal to
k3 x2 + k2 (x2 − x1 ) m1 x ¨1 = −m2 x ¨2
b−
b+
√
The special case; if k1 = k2 =k3 =k and m1 =m2 =m3 , then w1 and w2 is respectively equal to r
k m
(12)
3k m
(13)
w1 = x1 = x1,0 e
iwt
(3) r
and
w2 = x2 = x2,0 eiwt
(4)
Then if these equations are put into general formula, k2 (x2,0 − x1,0 ) − k1 x1,0 = −m1 x1,0 w2
(5)
k3 x2,0 + k2 (x2,0 − x1,0 ) = m2 x2,0 w2
(6)
w1 is the frequence which the two masses are in phase ( symmetrical mode) and w2 is the frequence that these masses are in out of phase (antisymmetrical mode). See fig. 2 and fig. 3
and
will obtained. The ratio of these equations are equal to
FIG. 2.
2
FIG. 3.
I.
SOLVING TWO COUPLED HARMONIC OSCILLATOR PROBLEM
Starting with te idea of a couple of masses , m1 and m2 , connected together with spring constants k1 , k2 and k3 . Two springs,k1 and k3 , connected to the stationary walls and the third one, k2 , connected between two masses. Equilibrium of the two masses are located in fig 1 at their centers.
k1 + k2 − m1 w2 k2 = (7) 2 k3 + k2 − m2 w k2 After the necessary simplification the equation goes to m1 m2 w4 − w2 (m2 k1 +m2 k2 + m1 k2 + m1 k3 ) + k3 k1 + k3 k2 + k2 k1 = 0
(8) (9)
If the coefficents are subsituted with a,b,c then; m1 .m2 = a m2 .k1 + m2 .k2 + m1 .k2 + m1 .k2 + m1 .k3 = b FIG. 1.
k3 .k1 + k3 .k2 + k2 .k1 = c The idea is these masses are going to stretch x1 and x2 from their equilibrium position. The forces on masses are going to be related to x1 and x2 . The force on the m1 , F1 , is equal to
Having the equations been solved s w1 =
k2 (x2 − x1 ) − k1 x1 = m1 x ¨1
(1)
s w1 = (2)
The resulting motion has an oscillatory behaviour, so the equations of x1 and x2 are
√
b2 − 4ac 2a
(10)
b2 − 4ac 2a
(11)
and
and on the m2, F2, is equal to
k3 x2 + k2 (x2 − x1 ) m1 x ¨1 = −m2 x ¨2
b−
b+
√
The special case; if k1 = k2 =k3 =k and m1 =m2 =m3 , then w1 and w2 is respectively equal to r
k m
(12)
3k m
(13)
w1 = x1 = x1,0 e
iwt
(3) r
and
w2 = x2 = x2,0 eiwt
(4)
Then if these equations are put into general formula, k2 (x2,0 − x1,0 ) − k1 x1,0 = −m1 x1,0 w2
(5)
k3 x2,0 + k2 (x2,0 − x1,0 ) = m2 x2,0 w2
(6)
w1 is the frequence which the two masses are in phase ( symmetrical mode) and w2 is the frequence that these masses are in out of phase (antisymmetrical mode). See fig. 2 and fig. 3
and
will obtained. The ratio of these equations are equal to
FIG. 2.
2
FIG. 3.
