Spring Mass System With External Force Finding Particular Solution

Parameter in physics problems

In a real spring–mass system, the spring has a non-negligible mass ${\displaystyle m}$ . Since not all of the spring's length moves at the same velocity ${\displaystyle v}$ as the suspended mass ${\displaystyle M}$ , its kinetic energy is not equal to ${\displaystyle {\tfrac {1}{2}}mv^{2}}$ . As such, ${\displaystyle m}$ cannot be simply added to ${\displaystyle M}$ to determine the frequency of oscillation, and the effective mass of the spring is defined as the mass that needs to be added to ${\displaystyle M}$ to correctly predict the behavior of the system.

Ideal uniform spring [edit]

vertical spring-mass system

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 ${\displaystyle u}$ is the velocity of mass element:

${\displaystyle T=\int _{m}{\tfrac {1}{2}}u^{2}\,dm}$

Since the spring is uniform, ${\displaystyle dm=\left({\frac {dy}{L}}\right)m}$ , where ${\displaystyle L}$ is the length of the spring. Hence,

${\displaystyle T=\int _{0}^{L}{\tfrac {1}{2}}u^{2}\left({\frac {dy}{L}}\right)m\!}$

${\displaystyle ={\tfrac {1}{2}}{\frac {m}{L}}\int _{0}^{L}u^{2}\,dy}$

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. ${\displaystyle u={\frac {vy}{L}}}$ , from which it follows:

${\displaystyle T={\tfrac {1}{2}}{\frac {m}{L}}\int _{0}^{L}\left({\frac {vy}{L}}\right)^{2}\,dy}$

${\displaystyle ={\tfrac {1}{2}}{\frac {m}{L^{3}}}v^{2}\int _{0}^{L}y^{2}\,dy}$

${\displaystyle ={\tfrac {1}{2}}{\frac {m}{L^{3}}}v^{2}\left[{\frac {y^{3}}{3}}\right]_{0}^{L}}$

${\displaystyle ={\tfrac {1}{2}}{\frac {m}{3}}v^{2}}$

Comparing to the expected original kinetic energy formula ${\displaystyle {\tfrac {1}{2}}mv^{2},}$ 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 ${\displaystyle x}$ from the spring's unstretched position (ignoring constant potential terms and taking the upwards direction as positive):

${\displaystyle T}$ (Total energy of system)

${\displaystyle ={\tfrac {1}{2}}({\frac {m}{3}})\ v^{2}+{\tfrac {1}{2}}Mv^{2}+{\tfrac {1}{2}}kx^{2}-{\tfrac {1}{2}}mgx-Mgx}$

Note that ${\displaystyle g}$ here is the acceleration of gravity along the spring. By differentiation of the equation with respect to time, the equation of motion is:

${\displaystyle \left({\frac {-m}{3}}-M\right)\ a=kx-{\tfrac {1}{2}}mg-Mg}$

The equilibrium point ${\displaystyle x_{\mathrm {eq} }}$ can be found by letting the acceleration be zero:

${\displaystyle x_{\mathrm {eq} }={\frac {1}{k}}\left({\tfrac {1}{2}}mg+Mg\right)}$

Defining ${\displaystyle {\bar {x}}=x-x_{\mathrm {eq} }}$ , the equation of motion becomes:

${\displaystyle \left({\frac {m}{3}}+M\right)\ a=-k{\bar {x}}}$

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

${\displaystyle \tau =2\pi \left({\frac {M+m/3}{k}}\right)^{1/2}}$

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 ${\displaystyle 2\pi \left({\frac {m}{k}}\right)^{1/2}}$ to determine the period of oscillation.

General case [edit]

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 ${\displaystyle \rho (x)}$ along its length:

${\displaystyle T=\int _{m}{\tfrac {1}{2}}u^{2}\,dm}$

${\displaystyle =\int _{0}^{L}{\tfrac {1}{2}}u^{2}\rho (x)\,dx}$

${\displaystyle =\int _{0}^{L}{\tfrac {1}{2}}\left({\frac {vx}{L}}\right)^{2}\rho (x)\,dx}$

${\displaystyle ={\tfrac {1}{2}}\left[\int _{0}^{L}{\frac {x^{2}}{L^{2}}}\rho (x)\,dx\right]v^{2}}$

So the effective mass of a spring is:

${\displaystyle m_{\mathrm {eff} }=\int _{0}^{L}{\frac {x^{2}}{L^{2}}}\rho (x)\,dx}$

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

Real spring [edit]

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 ${\displaystyle M/m}$ , the displacement is not so large as to cause elastic deformation. Jun-ichi Ueda and Yoshiro Sadamoto have found^[1] that as ${\displaystyle M/m}$ increases beyond 7, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value ${\displaystyle m/3}$ 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).

Application [edit]

The spring-mass system can be used in wide variety of applications. For example, it can be used to simulate the human tendon motions for computer graphics,^[2] and foot skin deformation.^[3]

See also [edit]

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

References [edit]

1. ^ Ueda, Jun-Ichi; Sadamoto, Yoshiro (1997). "A Measurement of the Effective Mass of Coil Springs". Journal of the Physical Society of Japan. 66 (2): 367–368. Bibcode:1997JPSJ...66..367U. doi:10.1143/JPSJ.66.367.
2. ^ Tang, Yuk-Ming; Hui, Kin-Chuen (2009-04-01). "Simulating tendon motion with axial mass–spring system". Computers & Graphics. 33 (2): 162–172. doi:10.1016/j.cag.2009.01.002. ISSN 0097-8493.
3. ^ Tang, Y. M.; Hui, K. C. (2007-07-01). "The effect of tendons on foot skin deformation". Computer-Aided Design. Human Modeling and Applications. 39 (7): 583–597. doi:10.1016/j.cad.2007.01.013. ISSN 0010-4485.

External links [edit]

• 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)

Spring Mass System With External Force Finding Particular Solution

Source: https://en.wikipedia.org/wiki/Effective_mass_(spring%E2%80%93mass_system)

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