4 Where Does the Wave Equation Come From?
23
with both constants of integration s o and φ we receive for φ = −1/2π the special
solution s = s o sin(ω t). Because of ˙
s = s o ω cos(ω t), the mass m reaches maximum
velocity v o = s o ω at time t = 0. For φ = 0, so that s = s o cos(ω t) it possesses
at t = 0 the maximum deflection out of the position of equilibrium x o . The mass
oscillates harmonically with the angular velocity ω = 2πν around the position of
equilibrium (oscillation period T = 1/ν). The function (14) is a solution for (13) if
and only if
ω =
D
m
.
(15)
The total energy U of the oscillation is equal to the maximum value of the kinetic
energy E kin =
1
2
m ˙
s 2 ,
U = max(E kin ) =
m
2
v
2
o =
m
2
s
2
o ω
2
=
1
2
s
2
o D.
(16)
We will now consider the harmonic oscillation (14). We count the number ν of times
the oscillation passes through the position of equilibrium per second, i.e. determine
the frequency ν. We can also measure the kinetic energy which is equal to the
total energy U when passing through the position of equilibrium. If for example
the oscillation is brought to a standstill at its position of equilibrium by friction, we
receive U out of the heat that was set free during the process.
We now observe a remarkable property. As long as we can only measure the angular velocity ω = 2πν =
√
D/m and the total energy U =
1
2
m s
2
o D, it is impossible
to determine separately the oscillating mass m and the force constant D. We cannot
decide if a mass of one tonne m = 1000 kg under the influence of the force constant
D = 10
7 Nm
−1 (Newton per metre), or if a tiny milligram with the mass m = 10
−6
kg oscillates under the influence of the force constant D = 10
−2 Nm
−1 . In both cases,
we get the same ω,
ω =
D
m
=
10 7
10 3
N
m
1
kg
=
10 −2
10 −6
N
m
1
kg
= 10
2
kg m
m s 2 kg
= 10
2 s
−1
,
in other words,
ν =
ω
2π
= 15, 9 Hz.
Even the measurement of the total energy cannot change this, because U also contains
the amplitude s o of an oscillation as a constant of integration. We can therefore say:
Both characteristics of a harmonic oscillation, its angular velocity ω and its total energy U ,
do not allow us to draw a conclusion about the mechanical state of the oscillating system.
23
with both constants of integration s o and φ we receive for φ = −1/2π the special
solution s = s o sin(ω t). Because of ˙
s = s o ω cos(ω t), the mass m reaches maximum
velocity v o = s o ω at time t = 0. For φ = 0, so that s = s o cos(ω t) it possesses
at t = 0 the maximum deflection out of the position of equilibrium x o . The mass
oscillates harmonically with the angular velocity ω = 2πν around the position of
equilibrium (oscillation period T = 1/ν). The function (14) is a solution for (13) if
and only if
ω =
D
m
.
(15)
The total energy U of the oscillation is equal to the maximum value of the kinetic
energy E kin =
1
2
m ˙
s 2 ,
U = max(E kin ) =
m
2
v
2
o =
m
2
s
2
o ω
2
=
1
2
s
2
o D.
(16)
We will now consider the harmonic oscillation (14). We count the number ν of times
the oscillation passes through the position of equilibrium per second, i.e. determine
the frequency ν. We can also measure the kinetic energy which is equal to the
total energy U when passing through the position of equilibrium. If for example
the oscillation is brought to a standstill at its position of equilibrium by friction, we
receive U out of the heat that was set free during the process.
We now observe a remarkable property. As long as we can only measure the angular velocity ω = 2πν =
√
D/m and the total energy U =
1
2
m s
2
o D, it is impossible
to determine separately the oscillating mass m and the force constant D. We cannot
decide if a mass of one tonne m = 1000 kg under the influence of the force constant
D = 10
7 Nm
−1 (Newton per metre), or if a tiny milligram with the mass m = 10
−6
kg oscillates under the influence of the force constant D = 10
−2 Nm
−1 . In both cases,
we get the same ω,
ω =
D
m
=
10 7
10 3
N
m
1
kg
=
10 −2
10 −6
N
m
1
kg
= 10
2
kg m
m s 2 kg
= 10
2 s
−1
,
in other words,
ν =
ω
2π
= 15, 9 Hz.
Even the measurement of the total energy cannot change this, because U also contains
the amplitude s o of an oscillation as a constant of integration. We can therefore say:
Both characteristics of a harmonic oscillation, its angular velocity ω and its total energy U ,
do not allow us to draw a conclusion about the mechanical state of the oscillating system.