m
v
M
M
1 | |
2 |
mv2
m
M
meff
M
The effective mass of the spring in a spring-mass system when using a heavy spring (non-ideal) of uniform linear density is
1 | |
3 |
The effective mass of the spring can be determined by finding its kinetic energy. For a differential mass element of the spring
dm
s
u(s)
dK=
1 | |
2 |
dmu2
In order to find the spring's total kinetic energy, it requires adding all the mass elements' kinetic energy, and requires the following integral:
K
=\int\limits | ||||
|
u2 dm
If one assumes a homogeneous stretching, the spring's mass distribution is uniform,
dm= | m |
y |
ds
y
K=
| ||||
\int | ||||
0 |
| ||||
u |
\right)ds
= | 1 |
2 |
m | |
y |
y | |
\int | |
0 |
u2ds
The velocity of each mass element of the spring is directly proportional to length from the position where it is attached (if near to the block then more velocity, and if near to the ceiling then less velocity), i.e.
u(s)= | s |
y |
v
K=
1 | |
2 |
m | |
y |
| ||||
\int | ||||
0 |
v\right)2ds
= | 1 |
2 |
m | |
y3 |
y | |
v | |
0 |
s2ds
= | 1 |
2 |
m | |
y3 |
| ||||
v |
y | |
\right] | |
0 |
= | 1 |
2 |
m | |
3 |
v2
Comparing to the expected original kinetic energy formula
1 | |
2 |
mv2,
m | |
3 |
To find the gravitational potential energy of the spring, one follows a similar procedure:
U=\int\limitsspringdmgs
=
y | |
\int | |
0 |
m | |
y |
gs ds
=mg
1 | |
y |
y | |
\int | |
0 |
s ds
=mg
1 | \left[ | |
y |
s2 | |
2 |
y | |
\right] | |
0 |
=
m | |
2 |
gy
Using this result, the total energy of system can be written in terms of the displacement
y
y=0
E=
1 | |
2 |
m | |
3 |
v2+
1 | |
2 |
Mv2+
1 | |
2 |
ky2+
1 | |
2 |
mgy+Mgy
=
1 | \left(M+ | |
2 |
m | |
3 |
\right)v2+
1 | |
2 |
ky2+\left(M+
m | |
2 |
\right)gy
Note that
g
\left(M+
m | |
3 |
\right) a=-ky-\left(M+
m | |
2 |
\right)g
The equilibrium point
yeq
yeq=-
| |||||
k |
Defining
\bar{y}=y-yeq
\left(M+
m | |
3 |
\right) a=-k\bar{y}=-k(y-yeq)
This is the equation for a simple harmonic oscillator with angular frequency:
\omega=\sqrt{
k | |||||
|
}
T=2\pi\sqrt{
| ||||||
k |
}
m | |
M |
\to0
So the effective mass of the spring added to the mass of the load gives us the "effective total mass" of the system that must be used in the standard formula
2\pi\sqrt{ | m |
k |
Finally, the solution to the initial value problem:
\left\{\begin{matrix}\ddot{y}&=&
2(y-y | ||
-\omega | \ | |
eq) |
y |
(0)&=&
y |
0\ y(0)&=&y0\end{matrix}\right.
y(t)=(y0-yeq)\cos(\omegat)+
| |||||
\omega |
\sin(\omegat)+yeq
As seen above, the effective mass of a spring does not depend upon "external" factors such as the acceleration of gravity along it. In fact, for a non-uniform heavy spring, the effective mass solely depends on its linear density
λ(s)
K=
\int\limits | ||||
|
u2 dm
=
| ||||
\int | ||||
0 |
u2(s)λ(s) ds
=
| |||||
\int | \left( | ||||
0 |
s | |
y |
v\right)2λ(s) ds
=
1 | |
2 |
\left(
y | |
\int | |
0 |
s2 | |
y2 |
λ(s) ds\right)v2
So the effective mass of a spring is:
meff=
y | |
\int | |
0 |
s2 | |
y2 |
λ(s)ds
This result also shows that
meff\leqslantm
meff=m
Three special cases can be considered:
λ(s)=0
meff=m=0
λ(s)= | m |
y |
m | ||||
|
λ(s)=m\delta(s-y)
\delta(x)
s=y
meff=m
To find the corresponding Lagrangian, one must find beforehand the potential gravitational energy of the spring:
U=\int\limitsspringdmgs
=
y | |
\int | |
0 |
λ(s)gs ds
=\left(
y | |
\int | |
0 |
s | |
y |
λ(s) ds\right)gy
Due to the monotonicity of the integral, it follows that:
0\leqslantmeff=
y | |
\int | |
0 |
s2 | |
y2 |
λ(s) ds\leqslant
y | |
\int | |
0 |
s | |
y |
λ(s) ds\leqslant
y | |
\int | |
0 |
λ(s) ds=m
With the Lagrangian being:
|
=
1 | |
2 |
\left(
y | |
M+\int | |
0 |
s2 | |
y2 |
λ(s) ds\right)
y |
2-
1 | |
2 |
ky2-\left(
y | |
M+\int | |
0 |
s | |
y |
λ(s) ds\right)gy
The above calculations assume that the stiffness coefficient of the spring does not depend on its length. However, this is not the case for real springs. For small values of
M | |
m |
M | |
m |
\ll1
m | ||||
|
m
M | |
m |
7
m | |
3 |
M | |
m |
≈ 38
Consider the pendulum differential equation:
2\sin\theta=0 | |
\ddot{\theta}+{\omega | |
0} |
Where
\omega0
2 | |
{\omega | |
0} |
g | |
\ell |
mgrCM | |
IO |
\ell
m
rCM
O
IO
Consider a system made of a homogeneous rod swinging from one end, and having attached bob at the other end. Let
\ell
mrod
mbob
λ(s)=
m | ||||
|
\delta( ⋅ )
mbob+mrod
rCM
(mbob+mrod)rCM=mbob\ell+m
|
\ell | |
\int | |
0 |
λ(s)(s-rCM) ds=0
r | |||||||||||||
|
\ell
IO=m
| ||||
bob\ell |
2 | |
m | |
rod\ell |
IO=\int
\ell | |
0 |
λ(s)s2 ds
2 | |
{\omega | |
0} |
=
mgrCM | = | |
IO |
| ||||||||||||||||||
|
=
g | |
\ell |
| ||||||||
|
=
g | |
\ell |
| ||||
|
Notice how the final expression is not a function on both the mass of the bob,
mbob
mrod
mrod | |
mbob |
mrod | |
mbob |
\ll1
| ||||
|
≈ 1+
1 | |
6 |
mrod | |
mbob |
-
1 | \left( | |
18 |
mrod | |
mbob |
\right)2+ …
Let's compare both results:
2 | |
{\omega | |
0} |
=\underbrace{
k | |
mbob |
2 | |
{\omega | |
0} |
=\underbrace{
g | |
\ell |