Effective mass (spring–mass system)

In a real spring–mass system, the spring has a non-negligible mass $m$. Since not all of the spring's length moves at the same velocity $v$ as the suspended mass $M$, its kinetic energy is not equal to $\tfrac m v^2$. As such, $m$ cannot be simply added to $M$ to determine the frequency of oscillation, and the effective mass of the spring is defined as the mass that needs to be added to $M$ to correctly predict the behavior of the system.

Ideal uniform spring

The effective mass of the spring in a spring-mass system when using an ideal spring 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. This requires adding all the mass elements' kinetic energy, and requires the following integral, where $u$ is the velocity of mass element:

:$K =\int_m\tfracu^2\,dm$

Since the spring is uniform, $dm=\left(\frac\right)m$, where $L$ is the length of the spring at the time of measuring the speed. Hence,

:$K = \int_0^L\tfracu^2\left(\frac\right)m\!$
::$=\tfrac\frac\int_0^L u^2\,dy$

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. $u=\frac$, from which it follows:

:$K =\tfrac\frac\int_0^L\left(\frac\right)^2\,dy$

:$=\tfrac\fracv^2\int_0^L y^2\,dy$

:=\tfrac\fracv^2\left frac\right0^L

:$=\tfrac\fracv^2$

Comparing to the expected original kinetic energy formula $\tfracmv^2,$ the effective mass of spring in this case is ''m''/3. Using this result, the total energy of system can be written in terms of the displacement $x$ from the spring's unstretched position (ignoring constant potential terms and taking the upwards direction as positive):

: $T$(Total energy of system)
::$= \tfrac(\frac)\ v^2 + \tfracM v^2 + \tfrac k x^2 - \tfracm g x - M g x$

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:

:$\left( \frac-M \right) \ a = kx -\tfrac mg - Mg$

The equilibrium point $x_$ can be found by letting the acceleration be zero:

:$x_ = \frac\left(\tfracmg + Mg \right)$

Defining $\bar = x - x_$, the equation of motion becomes:

:$\left( \frac+M \right) \ a = -k\bar$

This is the equation for a simple harmonic oscillator with period:

:$\tau = 2 \pi \left( \frac \right)^$

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 $2 \pi\sqrt$ to determine the period of oscillation.

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 spring, the effective mass solely depends on its linear density $\rho(x)$ along its length:

::$K = \int_m\tfracu^2\,dm$
:::$= \int_0^L\tfracu^2 \rho(x) \,dx$
:::$= \int_0^L\tfrac\left(\frac \right)^2 \rho(x) \,dx$
::: = \tfrac \left \int_0^L \frac \rho(x) \,dx \rightv^2

So the effective mass of a spring is:

:$m_ = \int_0^L \frac \rho(x) \,dx$

This result also shows that $m_ \le m$, with $m_ = m$ occurring in the case of an unphysical spring whose mass is located purely at the end farthest from the support.

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. Jun-ichi Ueda and Yoshiro Sadamoto have found 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. 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).

See also

* Simple harmonic motion (SHM) examples.
*Reduced mass

References

External links

*http://tw.knowledge.yahoo.com/question/question?qid=1405121418180
*http://tw.knowledge.yahoo.com/question/question?qid=1509031308350
*https://web.archive.org/web/20110929231207/http://hk.knowledge.yahoo.com/question/article?qid=6908120700201
*https://web.archive.org/web/20080201235717/http://www.goiit.com/posts/list/mechanics-effective-mass-of-spring-40942.htm
http://www.juen.ac.jp/scien/sadamoto_base/spring.html
*"The Effective Mass of an Oscillating Spring" Am. J. Phys., 38, 98 (1970)
*"Effective Mass of an Oscillating Spring" The Physics Teacher, 45, 100 (2007)

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Mechanical vibrations
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