Effective mass (spring–mass system) explained

m

. Since not all of the spring's length moves at the same velocity

v

as the suspended mass

M

(for example the point completely opposed to the mass

M

, at the other end of the spring, is not moving at all), its kinetic energy is not equal to
1
2

mv2

. As such,

m

cannot be simply added to

M

to determine the frequency of oscillation, and the effective mass of the spring,

meff

, is defined as the mass that needs to be added to

M

to correctly predict the behavior of the system.

Uniform spring (homogeneous)

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
of the mass of the spring and is independent of the direction of the spring-mass system (i.e., horizontal, vertical, and oblique systems all have the same effective mass). This is because external acceleration does not affect the period of motion around the equilibrium point.

The effective mass of the spring can be determined by finding its kinetic energy. For a differential mass element of the spring

dm

at a position

s

(dummy variable) moving with a speed

u(s)

, its kinetic energy is:

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
spring1
2

u2 dm

If one assumes a homogeneous stretching, the spring's mass distribution is uniform,

dm=m
y

ds

, where

y

is the length of the spring at the time of measuring the speed. Hence,

K=

y1
2
\int
0
2\left(m
y
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

, from which it follows:

K=

1
2
m
y
y\left(s
y
\int
0

v\right)2ds

=1
2
m
y3
y
v
0

s2ds

=1
2
m
y3
2\left[s3
3
v
y
\right]
0
=1
2
m
3

v2

Comparing to the expected original kinetic energy formula

1
2

mv2,

the effective mass of spring in this case is
m
3
. This result is known as Rayleigh's value, after Lord Rayleigh.

To find the gravitational potential energy of the spring, one follows a similar procedure:

U=\int\limitsspringdmgs

=

y
\int
0
m
y

gsds

=mg

1
y
y
\int
0

sds

=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

from the spring's unstretched position (taking the upwards direction as positive, ignoring constant potential terms and setting the origin of potential energy at

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

here is the acceleration of gravity along the spring. By differentiation of the equation with respect to time, the equation of motion is:

\left(M+

m
3

\right)a=-ky-\left(M+

m
2

\right)g

The equilibrium point

yeq

can be found by letting the acceleration be zero:

yeq=-

\left(M+m\right)g
2
k

Defining

\bar{y}=y-yeq

, the equation of motion becomes:

\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
M+
m
3

}

Thus, it has a smaller angular frequency than in the ideal spring. Also, its period is given by:

T=2\pi\sqrt{

M+
m
3
k

}

Which is bigger than the ideal spring. Both formulae reduce to the ideal case in the limit
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
} to determine the period of oscillation.

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.

Is given by:

y(t)=(y0-yeq)\cos(\omegat)+

y0
\omega

\sin(\omegat)+yeq

Which is a simple harmonic motion.

General case

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)

along its length:

K=

\int\limits
spring1
2

u2 dm

=

y1
2
\int
0

u2(s)λ(s)ds

=

y1
2
\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

, with

meff=m

occurring in the case of an unphysical spring whose mass is located purely at the end farthest from the support.

Three special cases can be considered:

λ(s)=0

is the idealised case where the spring has no mass, and

meff=m=0

.
λ(s)=m
y

is the homogeneous case (uniform spring) where Rayleigh's value appears in the equation, i.e.,
m
eff=m
3
.

λ(s)=m\delta(s-y)

, where

\delta(x)

is Dirac delta function, is the extreme case when all the mass is located at

s=y

, resulting in

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)gsds

=\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:

l{L}(y,y
)

=

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

Real spring

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
, the displacement is not so large as to cause elastic deformation. In fact for
M
m

\ll1

, the effective mass is
m
eff=4
\pi2

m

. Jun-ichi Ueda and Yoshiro Sadamoto have found[1] that as
M
m
increases beyond

7

, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value
m
3
and eventually reaches negative values at about
M
m

38

. This unexpected behavior of the effective mass can be explained in terms of the elastic after-effect (which is the spring's not returning to its original length after the load is removed).

Comparision with pendulum

Consider the pendulum differential equation:

2\sin\theta=0
\ddot{\theta}+{\omega
0}

Where

\omega0

is the natural frequency of oscillations (and the angular frequency for small oscillations). The parameter
2
{\omega
0}
stands for
g
\ell
in an ideal pendulum, and
mgrCM
IO
in a compound pendulum, where

\ell

is the length of the pendulum,

m

is the total mass of the system,

rCM

is the distance from the pivot point

O

(the point the pendulum is suspended from) to the pendulum's centre-of-mass, and

IO

is the moment of inertia of the system with respect to an axis that goes through the pivot.

Consider a system made of a homogeneous rod swinging from one end, and having attached bob at the other end. Let

\ell

be the length of the rod,

mrod

the mass of the rod, and

mbob

the mass of the bob, thus the linear density is given by

λ(s)=

m
bob\delta(s-\ell)+mrod
\ell
, with

\delta()

Dirac's delta function. The total mass of the system is

mbob+mrod

. To find out

rCM

one must solve

(mbob+mrod)rCM=mbob\ell+m

rod\ell
2
by definition of centre-of-mass (this would be an integral equation in the general case,
\ell
\int
0

λ(s)(s-rCM)ds=0

,but it simplifies to this in the homogeneous case), whose solution is give by
r
CM=
mmrod
bob+1
2
mbob+mrod

\ell

. The moment of inertia of the system is the sum of the two moments of inertia,

IO=m

2+1
3
bob\ell
2
m
rod\ell
(once again in the general case the integral equation would be

IO=\int

\ell
0

λ(s)s2 ds

). Thus the expression can be simplified:
2
{\omega
0}

=

mgrCM=
IO
\left(m
rod\ell
2
\right)g
bob\ell+m
2+1
3
m
2
m
rod\ell
bob\ell

=

g
\ell
m
bob+mrod
2
m
bob+mrod
3

=

g
\ell
1+mrod
2mbob
1+mrod
3mbob

Notice how the final expression is not a function on both the mass of the bob,

mbob

, and the mass of the rod,

mrod

, but only on their ratio,
mrod
mbob
. Also notice that initially it has the same structure as the spring-mass system: the product of the ideal case and a correction (with Rayleigh's value). Notice that for
mrod
mbob

\ll1

, the last correction term can be approximated by:
1+mrod
2mbob
1+mrod
3mbob

1+

1
6
mrod
mbob

-

1\left(
18
mrod
mbob

\right)2+

Let's compare both results:

2
{\omega
0}

=\underbrace{

k
mbob
}_\text \underbrace_
2
{\omega
0}

=\underbrace{

g
\ell
}_\text \underbrace_ \underbrace_

See also

External links

Notes and References

  1. 10.1143/JPSJ.66.367. A Measurement of the Effective Mass of Coil Springs. Journal of the Physical Society of Japan. 66. 2. 367–368. 1997. Ueda. Jun-Ichi. Sadamoto. Yoshiro. 1997JPSJ...66..367U.