Effective mass (spring–mass system)

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

$${\displaystyle \begin{array}{cc}K& =\underset{\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}}{\int }\frac{1}{2}{u}^2\mathrm{d}m\\ & ={\int }_0^y\frac{1}{2}{u}^2(s)\lambda (s)\mathrm{d}s\\ & ={\int }_0^y\frac{1}{2}{\left(\frac{s}{y}v\right)}^2\lambda (s)\mathrm{d}s\\ & =\frac{1}{2}\left({\int }_0^y\frac{s^2}{y^2}\lambda (s)\mathrm{d}s\right)v^2\end{array}}$$

So the effective mass of a spring is:

$${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}={\int }_0^y\frac{s^2}{y^2}\lambda (s)\mathrm{d}s}$$

This result also shows that ${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}⩽m}$, with ${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}=m}$ 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 ${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}=m=0}$.
• ${\displaystyle \lambda (s)=\frac{m}{y}}$ is the homogeneous case (uniform spring) where Rayleigh's value appears in the equation, i.e., ${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}=\frac{m}{3}}$.
• ${\displaystyle \lambda (s)=m\delta (s-y)}$, where ${\displaystyle \delta (x)}$ is Dirac delta function, is the extreme case when all the mass is located at ${\displaystyle s=y}$, resulting in ${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}=m}$.

To find the corresponding Lagrangian, one must find beforehand the potential gravitational energy of the spring: $${\displaystyle \begin{array}{cc}U& =\underset{\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}}{\int }\mathrm{d}mgs\\ & ={\int }_0^y\lambda (s)gs\mathrm{d}s\\ & =\left({\int }_0^y\frac{s}{y}\lambda (s)\mathrm{d}s\right)gy\end{array}}$$

Due to the monotonicity of the integral, it follows that: $${\displaystyle 0⩽m_{\mathrm{e}\mathrm{f}\mathrm{f}}={\int }_0^y\frac{s^2}{y^2}\lambda (s)\mathrm{d}s⩽{\int }_0^y\frac{s}{y}\lambda (s)\mathrm{d}s⩽{\int }_0^y\lambda (s)\mathrm{d}s=m}$$

With the Lagrangian being: $${\displaystyle \mathcal{L}(y,\dot{y})=\frac{1}{2}\left(M+{\int }_0^y\frac{s^2}{y^2}\lambda (s)\mathrm{d}s\right){\dot{y}}^2-\frac{1}{2}k{y}^2-\left(M+{\int }_0^y\frac{s}{y}\lambda (s)\mathrm{d}s\right)gy}$$

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 ${\displaystyle \frac{M}{m}}$, the displacement is not so large as to cause elastic deformation. In fact for ${\displaystyle \frac{M}{m}\ll 1}$, the effective mass is ${\displaystyle m_{\mathrm{e}\mathrm{f}\mathrm{f}}=\frac{4}{{\pi }^2}m}$. Jun-ichi Ueda and Yoshiro Sadamoto have found 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).

Comparison with pendulum

Consider the pendulum differential equation:

$${\displaystyle \ddot{\theta }+{{\omega }_0}^2\sin \theta =0}$$

Where ${\displaystyle {\omega }_0}$ is the natural frequency of oscillations (and the angular frequency for small oscillations). The parameter ${\displaystyle {{\omega }_0}^2}$ stands for ${\displaystyle \frac{g}{\ell }}$ in an ideal pendulum, and ${\displaystyle \frac{mg{r}_{\mathrm{C}\mathrm{M}}}{I_O}}$ in a compound pendulum, where ${\displaystyle \ell }$ is the length of the pendulum, ${\displaystyle m}$ is the total mass of the system, ${\displaystyle r_{\mathrm{C}\mathrm{M}}}$ is the distance from the pivot point ${\displaystyle O}$ (the point the pendulum is suspended from) to the pendulum's centre-of-mass, and ${\displaystyle I_O}$ 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 ${\displaystyle \ell }$ be the length of the rod, ${\displaystyle m_{\mathrm{r}\mathrm{o}\mathrm{d}}}$ the mass of the rod, and ${\displaystyle m_{\mathrm{b}\mathrm{o}\mathrm{b}}}$ the mass of the bob, thus the linear density is given by ${\displaystyle \lambda (s)=m_{\mathrm{b}\mathrm{o}\mathrm{b}}\delta (s-\ell )+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{\ell }}$, with ${\displaystyle \delta (\cdot )}$ Dirac's delta function. The total mass of the system is ${\displaystyle m_{\mathrm{b}\mathrm{o}\mathrm{b}}+m_{\mathrm{r}\mathrm{o}\mathrm{d}}}$. To find out ${\displaystyle r_{\mathrm{C}\mathrm{M}}}$ one must solve ${\displaystyle (m_{\mathrm{b}\mathrm{o}\mathrm{b}}+m_{\mathrm{r}\mathrm{o}\mathrm{d}})r_{\mathrm{C}\mathrm{M}}=m_{\mathrm{b}\mathrm{o}\mathrm{b}}\ell +m_{\mathrm{r}\mathrm{o}\mathrm{d}}\frac{\ell }{2}}$ by definition of centre-of-mass (this would be an integral equation in the general case, ${\displaystyle {\int }_0^{\ell }\lambda (s)(s-r_{\mathrm{C}\mathrm{M}})\mathrm{d}s=0}$, but it simplifies to this in the homogeneous case), whose solution is give by ${\displaystyle r_{\mathrm{C}\mathrm{M}}=\frac{m_{\mathrm{b}\mathrm{o}\mathrm{b}}+\frac{1}{2}{m}_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}+m_{\mathrm{r}\mathrm{o}\mathrm{d}}}\ell }$. The moment of inertia of the system is the sum of the two moments of inertia, ${\displaystyle I_O=m_{\mathrm{b}\mathrm{o}\mathrm{b}}{\ell }^2+\frac{1}{3}{m}_{\mathrm{r}\mathrm{o}\mathrm{d}}{\ell }^2}$ (once again in the general case the integral equation would be ${\displaystyle I_O={\int }_0^{\ell }\lambda (s)s^2\mathrm{d}s}$). Thus the expression can be simplified:

$${\displaystyle {{\omega }_0}^2=\frac{mg{r}_{\mathrm{C}\mathrm{M}}}{I_O}=\frac{\left(m_{\mathrm{b}\mathrm{o}\mathrm{b}}\ell +m_{\mathrm{r}\mathrm{o}\mathrm{d}}\frac{\ell }{2}\right)g}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}{\ell }^2+\frac{1}{3}{m}_{\mathrm{r}\mathrm{o}\mathrm{d}}{\ell }^2}=\frac{g}{\ell }\frac{m_{\mathrm{b}\mathrm{o}\mathrm{b}}+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{2}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{3}}=\frac{g}{\ell }\frac{1+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{2m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}{1+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{3m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}}$$

Notice how the final expression is not a function on both the mass of the bob, ${\displaystyle m_{\mathrm{b}\mathrm{o}\mathrm{b}}}$, and the mass of the rod, ${\displaystyle m_{\mathrm{r}\mathrm{o}\mathrm{d}}}$, but only on their ratio,${\displaystyle \frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}$. 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 ${\displaystyle \frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}\ll 1}$, the last correction term can be approximated by:

$${\displaystyle \frac{1+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{2m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}{1+\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{3m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}\approx 1+\frac{1}{6}\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}-\frac{1}{18}{\left(\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}\right)}^2+\cdots }$$

Let's compare both results:

• For the spring–mass system: $${\displaystyle {{\omega }_0}^2=\underset{\text{Ideal case}}{\underbrace{\frac{k}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}}\underset{\text{Correction with Rayleigh's value}}{\underbrace{\frac{1}{1+\frac{1}{3}\frac{m_{\mathrm{s}\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}}}}$$
• For the pendulum: $${\displaystyle {{\omega }_0}^2=\underset{\text{Ideal case}}{\underbrace{\frac{g}{\ell }}}\underset{\text{Correction with Rayleigh's value}}{\underbrace{\frac{1}{1+\frac{1}{3}\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}}}}\underset{\text{Rest of the correction}}{\underbrace{\left(1+\frac{1}{2}\frac{m_{\mathrm{r}\mathrm{o}\mathrm{d}}}{m_{\mathrm{b}\mathrm{o}\mathrm{b}}}\right)}}}$$