Effective mass (spring–mass system)

vertical spring-mass system

The effective mass of the spring in a spring-mass system when using a heavy spring (non-ideal) of uniform linear density is ${\displaystyle {\frac {1}{3}}}$ 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 ${\displaystyle \mathrm {d} m}$ at a position ${\displaystyle s}$ (dummy variable) moving with a speed ${\displaystyle u(s)}$, its kinetic energy is:

${\displaystyle \mathrm {d} K={\frac {1}{2}}\mathrm {d} m\,u^{2}}$

In order to find the spring's total kinetic energy, it requires adding all the mass elements' kinetic energy, and requires the following integral:

${\displaystyle K=\int \limits _{\mathrm {spring} }{\frac {1}{2}}u^{2}\;\mathrm {d} m}$

If one assumes a homogeneous stretching, the spring's mass distribution is uniform, ${\displaystyle \mathrm {d} m={\frac {m}{y}}\mathrm {d} s}$, where ${\displaystyle y}$ is the length of the spring at the time of measuring the speed. Hence,

${\displaystyle K=\int _{0}^{y}{\frac {1}{2}}u^{2}\left({\frac {m}{y}}\right)\mathrm {d} s}$

${\displaystyle ={\frac {1}{2}}{\frac {m}{y}}\int _{0}^{y}u^{2}\mathrm {d} s}$

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. ${\displaystyle u(s)={\frac {s}{y}}v}$, from which it follows:

${\displaystyle K={\frac {1}{2}}{\frac {m}{y}}\int _{0}^{y}\left({\frac {s}{y}}v\right)^{2}\mathrm {d} s}$

${\displaystyle ={\frac {1}{2}}{\frac {m}{y^{3}}}v^{2}\int _{0}^{y}s^{2}\,\mathrm {d} s}$

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

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

Comparing to the expected original kinetic energy formula ${\displaystyle {\frac {1}{2}}mv^{2},}$ the effective mass of spring in this case is ${\displaystyle {\frac {m}{3}}}$. 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:

${\displaystyle U=\int \limits _{\mathrm {spring} }\mathrm {d} m\,gs}$

${\displaystyle =\int _{0}^{y}{\frac {m}{y}}gs\;\mathrm {d} s}$

${\displaystyle =mg\,{\frac {1}{y}}\int _{0}^{y}s\;\mathrm {d} s}$

${\displaystyle =mg\,{\frac {1}{y}}\left[{\frac {s^{2}}{2}}\right]_{0}^{y}}$

${\displaystyle ={\frac {m}{2}}gy}$

Using this result, the total energy of system can be written in terms of the displacement ${\displaystyle y}$ from the spring's unstretched position (taking the upwards direction as positive, ignoring constant potential terms and setting the origin of potential energy at ${\displaystyle y=0}$):

${\displaystyle E={\frac {1}{2}}{\frac {m}{3}}v^{2}+{\frac {1}{2}}Mv^{2}+{\frac {1}{2}}ky^{2}+{\frac {1}{2}}mgy+Mgy}$

${\displaystyle ={\frac {1}{2}}\left(M+{\frac {m}{3}}\right)v^{2}+{\frac {1}{2}}ky^{2}+\left(M+{\frac {m}{2}}\right)gy}$

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:

${\displaystyle \left(M+{\frac {m}{3}}\right)\ a=-ky-\left(M+{\frac {m}{2}}\right)g}$

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

${\displaystyle y_{\mathrm {eq} }=-{\frac {\left(M+{\frac {m}{2}}\right)g}{k}}}$

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

${\displaystyle \left(M+{\frac {m}{3}}\right)\ a=-k{\bar {y}}=-k(y-y_{\mathrm {eq} })}$

This is the equation for a simple harmonic oscillator with angular frequency:

${\displaystyle \omega ={\sqrt {{\frac {k}{M+{\frac {m}{3}}}}\,}}}$

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

${\displaystyle T=2\pi {\sqrt {{\frac {M+{\frac {m}{3}}}{k}}\,}}}$

Which is bigger than the ideal spring. Both formulae reduce to the ideal case in the limit ${\displaystyle {\frac {m}{M}}\to 0}$.

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 ${\displaystyle 2\pi {\sqrt {\frac {m}{k}}}}$ to determine the period of oscillation.

Finally, the solution to the initial value problem:

${\displaystyle \left\{{\begin{matrix}{\ddot {y}}&=&-\omega ^{2}(y-y_{\mathrm {eq} })\\{\dot {y}}(0)&=&{\dot {y}}_{0}\\y(0)&=&y_{0}\end{matrix}}\right.}$

Is given by:

${\displaystyle y(t)=(y_{0}-y_{\text{eq}})\cos(\omega t)+{\frac {{\dot {y}}_{0}}{\omega }}\sin(\omega t)+y_{\text{eq}}}$

Which is a simple harmonic motion.

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 ${\displaystyle \lambda (s)}$ along its length:

${\displaystyle K=\int \limits _{\mathrm {spring} }{\frac {1}{2}}u^{2}\;\mathrm {d} m}$

${\displaystyle =\int _{0}^{y}{\frac {1}{2}}u^{2}\!(s)\,\lambda (s)\;\mathrm {d} s}$

${\displaystyle =\int _{0}^{y}{\frac {1}{2}}\left({\frac {s}{y}}v\right)^{2}\lambda (s)\;\mathrm {d} s}$

${\displaystyle ={\frac {1}{2}}\left(\int _{0}^{y}{\frac {s^{2}}{y^{2}}}\lambda (s)\;\mathrm {d} s\right)v^{2}}$

So the effective mass of a spring is:

${\displaystyle m_{\mathrm {eff} }=\int _{0}^{y}{\frac {s^{2}}{y^{2}}}\lambda (s)\,\mathrm {d} s}$

This result also shows that ${\displaystyle m_{\mathrm {eff} }\leqslant m}$, 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.

Three special cases can be considered:

• ${\displaystyle \lambda (s)=0}$ is the idealised case where the spring has no mass, and ${\displaystyle m_{\mathrm {eff} }=m=0}$.
• ${\displaystyle \lambda (s)={\frac {m}{y}}}$ is the homogeneous case (uniform spring) where Rayleigh's value appears in the equation, i.e. , ${\displaystyle m_{\mathrm {eff} }={\frac {m}{3}}}$.
• ${\displaystyle \lambda (s)=m\,\delta (s-y)}$, where ${\displaystyle \delta (x)}$ is Dirac delta function, is the extreme case when all the mass is located at ${\displaystyle s=y}$, resulting in ${\displaystyle m_{\mathrm {eff} }=m}$.

To find the corresponding Lagrangian, one must find beforehand the potential gravitational energy of the spring:

${\displaystyle U=\int \limits _{\mathrm {spring} }\mathrm {d} m\,g\,s}$

${\displaystyle =\int _{0}^{y}\lambda (s)\,g\,s\;\mathrm {d} s}$

${\displaystyle =\left(\int _{0}^{y}{\frac {s}{y}}\lambda (s)\;\mathrm {d} s\right)g\,y}$

Due to the monotonicity of the integral, it follows that: ${\displaystyle 0\leqslant m_{\mathrm {eff} }=\int _{0}^{y}{\frac {s^{2}}{y^{2}}}\lambda (s)\;\mathrm {d} s\leqslant \int _{0}^{y}{\frac {s}{y}}\lambda (s)\;\mathrm {d} s\leqslant \int _{0}^{y}\lambda (s)\;\mathrm {d} s=m}$

With the Lagrangian being: ${\displaystyle {\mathcal {L}}(y,{\dot {y}})={\frac {1}{2}}\left(M+\int _{0}^{y}{\frac {s^{2}}{y^{2}}}\lambda (s)\;\mathrm {d} s\right){\dot {y}}^{2}-{\frac {1}{2}}ky^{2}-\left(M+\int _{0}^{y}{\frac {s}{y}}\lambda (s)\;\mathrm {d} s\right)g\,y}$

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 ${\displaystyle {\frac {M}{m}}}$, the displacement is not so large as to cause elastic deformation. In fact for ${\displaystyle {\frac {M}{m}}\ll 1}$, the effective mass is ${\displaystyle m_{\mathrm {eff} }={\frac {4}{\pi ^{2}}}m}$ . Jun-ichi Ueda and Yoshiro Sadamoto have found[1] that as ${\displaystyle {\frac {M}{m}}}$ increases beyond ${\displaystyle 7}$, the effective mass of a spring in a vertical spring-mass system becomes smaller than Rayleigh's value ${\displaystyle {\frac {m}{3}}}$ and eventually reaches negative values at about ${\displaystyle {\frac {M}{m}}\approx 38}$. 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).

• 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)