Effective mass (spring–mass system)

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 small element near 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:

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

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:

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

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.

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.

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).

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

• 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)