24
4 Where Does the Wave Equation Come From?
Fig. 4.3 Model of a harmonic oscillating mass m
We can use a single mass m suspended between two walls at a position of rest
x o by two springs, which possess the force constants D/2 as a model for harmonic
oscillation; see Fig. 4.3. We once again denote the deflection out of the position of
equilibrium as s = x − x o . Therefore, the left spring pulls the mass m with the same
force f l = −
1
2
D s to the left as the right spring pushes the mass to the right with the
force F r = −
1
2
D s. The total resulting force acting on the mass m is F = f l + f r =
−D s, and the equation for an oscillation (13) is valid.
We now consider an oscillating system made up out of two equal masses of the
size m/2 coupled together with the force constant (3/4)D and attached to the wall
with the force constant D/2; see Fig. 4.4.
The instantaneous positions of the first and second mass are x 1 and x 2 . Let x o1 and
x o2 be their positions of equilibrium, so that x 1 = x o1 + s 1 and x 2 = x o2 + s 2 with
the deflections s 1 and s 2 out of their positions of rest. Their velocities are therefore
v 1 = ˙
x 1 = ˙
s 1 , v 2 = ˙
x 2 = ˙
s 2 and the forces F 1 or F 2 act on both masses according to
F 1 = −
1
2
D (x 1 − x o1 ) +
3
4
D [(x 2 − x 1 ) − (x o2 − x o1 )] = −
5
4
D s 1 +
3
4
D s 2
and
F 2 = −
5
4
D s 2 +
3
4
D s 1 .
4 Where Does the Wave Equation Come From?
Fig. 4.3 Model of a harmonic oscillating mass m
We can use a single mass m suspended between two walls at a position of rest
x o by two springs, which possess the force constants D/2 as a model for harmonic
oscillation; see Fig. 4.3. We once again denote the deflection out of the position of
equilibrium as s = x − x o . Therefore, the left spring pulls the mass m with the same
force f l = −
1
2
D s to the left as the right spring pushes the mass to the right with the
force F r = −
1
2
D s. The total resulting force acting on the mass m is F = f l + f r =
−D s, and the equation for an oscillation (13) is valid.
We now consider an oscillating system made up out of two equal masses of the
size m/2 coupled together with the force constant (3/4)D and attached to the wall
with the force constant D/2; see Fig. 4.4.
The instantaneous positions of the first and second mass are x 1 and x 2 . Let x o1 and
x o2 be their positions of equilibrium, so that x 1 = x o1 + s 1 and x 2 = x o2 + s 2 with
the deflections s 1 and s 2 out of their positions of rest. Their velocities are therefore
v 1 = ˙
x 1 = ˙
s 1 , v 2 = ˙
x 2 = ˙
s 2 and the forces F 1 or F 2 act on both masses according to
F 1 = −
1
2
D (x 1 − x o1 ) +
3
4
D [(x 2 − x 1 ) − (x o2 − x o1 )] = −
5
4
D s 1 +
3
4
D s 2
and
F 2 = −
5
4
D s 2 +
3
4
D s 1 .