Spring-Mass Systems

Abstract

The single degree-of-freedom system subject to mass and base excitation is used to model an elastic system to determine the frequency-domain effects of squeeze film air damping and viscous fluid damping. This model is also used to determine the important response characteristics of electrostatic attraction and van der Waals forces, the maximum average power from piezoelectric and electromagnetic coupling, and to illustrate the fundamental working principle of an atomic force microscope. The two degree-of-freedom system is introduced to examine microelectromechanical filters, atomic force microscope specimen control devices, and as a means to increase the input to piezoelectric energy harvesters. An appendix gives the details of the derivation of a hydrodynamic function that expresses the effects of a viscous fluid on a vibrating cylinder.

Appendix 2.1 Forces on a Submerged Vibrating Cylinder

Consider a very long rigid solid circular cylinder of diameter b and length L that is immersed in an incompressible viscous fluid of density ρ _f and dynamic viscosity μ _f . It is assumed that the fluid can be modeled by the linearized momentum equations, which in terms of the scalar stream function $\psi = \psi \left(r,\theta ,t\right)$ is given by (Stokes 1901, pp. 38–47)

$$\nabla ^2 \left( {\nabla ^2 \psi - \frac{{\rho _f }}{{\mu _f }}\frac{{\partial \psi }}{{\partial t}}} \right) = 0$$

((2.247))

where

$$\nabla ^2 = \frac{{\partial ^2 }}{{\partial r^2 }} + \frac{1}{r}\frac{\partial }{{\partial r}} + \frac{1}{{r^2 }}\frac{{\partial ^2 }}{{\partial \theta ^2 }}.$$

((2.248))

A solution to Eq. (2.247) is $\psi = \psi _1 + \psi _2$, where ψ ₁ and ψ ₂, respectively, are solutions to

$$\begin{array}{rl} \nabla ^2 \psi _1 =& 0 \\ \nabla ^2 \psi _2 - \dfrac{{\rho _f }}{{\mu _f }}\dfrac{{\partial \psi _2 }}{{\partial t}} =& 0. \\ \end{array}$$

((2.249))

The velocities of the fluid are given by

$$u_r = \frac{1}{r}\frac{{\partial \psi }}{{\partial \theta }},\quad{\textrm{ }}u_\theta = - \frac{{\partial \psi }}{{\partial r}}.$$

((2.250))

We assume that the origin of the polar coordinate system is at the center of the cylinder, and that the cylinder is undergoing forced harmonic oscillations at frequency ω of the form

$$\begin{array}{rl} u_r \left(b/2,\theta ,t\right) =& u_o \cos \theta e^{j\omega t} \\ u_\theta \left(b/2,\theta ,t\right) =& - u_o \sin \theta e^{j\omega t} \\ \end{array}$$

((2.251))

where u _o is the magnitude of the velocity. It is further assumed that the surrounding fluid is infinite in extent, that is, as $r \to \infty$, $u_r \to 0$ and $u_\theta \to 0$.

In view of the boundary conditions given by Eq. (2.251), we assume solutions to Eq. (2.249) of the form

$$\psi _k = \Psi _k \left(r\right)\sin \theta e\,^{j\omega t} {\textrm{\quad }}k = 1,2.$$

((2.252))

Substituting Eq. (2.252) into Eq. (2.249) yields

$$\nabla _r^2 \Psi _1 = 0$$

((2.253))

and

$$\nabla _r^2 \Psi _2 - \frac{{j\omega \rho _f }}{{\mu _f }}\Psi _2 = 0$$

((2.254))

where

$$\nabla _r^2 = \frac{{\partial ^2 }}{{\partial r^2 }} + \frac{1}{r}\frac{\partial }{{\partial r}} - \frac{1}{{r^2 }}.$$

((2.255))

The solution to Eq. (2.253) is

$$\Psi _1 = \frac{{C_1 }}{r} + C_2 r.$$

((2.256))

To satisfy the requirements that as $r \to \infty$, $u_r \to 0$ and $u_\theta \to 0$, we must set C ₂ = 0. Then, Eq. (2.256) becomes

$$\Psi _1 = \frac{{C_1 }}{r}.$$

((2.257))

To determine the solution to Eq. (2.254), we rewrite it as

$$r^2 \frac{{\partial ^2 \Psi _2 }}{{\partial r^2 }} + r\frac{{\partial \Psi _2 }}{{\partial r}} - \left( {\left( {\frac{{2\lambda r}}{b}} \right)^2 + 1} \right)\Psi _2 = 0$$

((2.258))

where

$$\begin{array}{rl} \lambda = & \sqrt {j\,{\textrm{Re}} } \\ {\textrm{Re}} =& \dfrac{{\rho _f \omega b^2 }}{{4\mu _f }}. \\ \end{array}$$

((2.259))

and Re is the Reynolds number . The solution to Eq. (2.258) is

$$\Psi _2 = C_3 I_1 \left(2\lambda r/b\right) + C_4 K_1 \left(2\lambda r/b\right)$$

((2.260))

where I ₁(x) is the modified Bessel function of the first kind of order 1 and K ₁(x) is the modified Bessel function of the second kind of order 1. To satisfy the requirements that as $r \to \infty$, $u_r \to 0$ and $u_\theta \to 0$, we must set C ₃ = 0. Then, Eq. (2.260) becomes

$$\Psi _2 = C_4 K_1 \left(2\lambda r/b\right).$$

((2.261))

Thus, from Eqs. (2.257) and (2.261),

$$\begin{array}{rl} \psi = & \Psi \left(r\right)\sin \theta e\,^{j\omega t} = \left( {\Psi _1 + \Psi _2 } \right)\sin \theta e\,^{j\omega t} \\ {}=& \left( {\dfrac{{C_1 }}{r} + C_4 K_1 \left(2\lambda r/b\right)} \right)\sin \theta e\,^{j\omega t} . \\ \end{array}$$

((2.262))

Then, from Eq. (2.250) and (2.262),

$$\begin{array}{rl} u_r = & \displaystyle\frac{1}{r}\left( {\frac{{C_1 }}{r} + C_4 K_1 \left(2\lambda r/b\right)} \right)\cos \theta e^{j\omega t} \\ u_\theta = & \displaystyle\left( {\frac{{C_1 }}{{r^2 }} + \frac{{2\lambda C_4 }}{b}\left[ {\frac{b}{{2\lambda r}}K_1 \left(2\lambda r/b\right) + K_0 \left(2\lambda r/b\right)} \right]} \right)\sin \theta e^{j\omega t} . \\ \end{array}$$

((2.263))

The constants C ₁ and C ₄ are determined by substituting Eq. (2.263) into the boundary conditions given by Eq. (2.251). This substitution results in

$$\begin{array}{rl} u_o =&\displaystyle \frac{{4C_1 }}{{b^2 }} + \frac{2}{b}C_4 K_1 \left(\lambda \right) \\ - u_o = &\displaystyle\frac{{4C_1 }}{{b^2 }} + \frac{{2\lambda C_4 }}{b}\left[ {\frac{1}{\lambda }K_1 \left(\lambda \right) + K_0 \left(\lambda \right)} \right]. \\ \end{array}$$

((2.264))

Solving Eq. (2.264) for C ₁ and C ₄, we obtain

$$\begin{array}{rl} C_1 =&\displaystyle \frac{{u_o b^2 }}{4}\left( {\frac{2}{\lambda }\frac{{K_1 \left(\lambda \right)}}{{K_0 \left(\lambda \right)}} + 1} \right) \\ C_4 =&\displaystyle - \frac{{u_o b}}{{\lambda K_0 \left(\lambda \right)}}. \\ \end{array}$$

((2.265))

Then Eqs. (2.261) and (2.257), respectively, become

$$\Psi _2 = - u_o b\frac{{K_1 \left(2\lambda r/b\right)}}{{\lambda K_0 \left(\lambda \right)}}$$

((2.266))

and

$$\Psi _1 = \frac{{u_o b^2 }}{{4r}}\left( {\frac{2}{\lambda }\frac{{K_1 \left(\lambda \right)}}{{K_0 \left(\lambda \right)}} + 1} \right).$$

((2.267))

The force F acting on the cylinder of length L at $r = b/2$ is determined from

$$F = L\frac{b}{2}\int\limits_0^{2\pi } {\left[ {P_r \cos \theta - P_\theta \sin \theta } \right]d\theta } $$

((2.268))

where

$$\begin{array}{rl} P_r =&\displaystyle - \left( p \right)_{r = b/2} + 2\mu _f \left( {\frac{{\partial u_r }}{{\partial r}}} \right)_{r = b/2} \\ P_\theta = & \displaystyle\mu _f \left\{ {\left( {\frac{1}{r}\frac{{\partial u_r }}{{\partial \theta }}} \right)_{r = b/2} + \left( {\frac{{\partial u_\theta }}{{\partial r}}} \right)_{r = b/2} - \left( {\frac{{u_\theta }}{r}} \right)_{r = b/2} } \right\} \\ \end{array}$$

((2.269))

and p is the pressure in the fluid. We now use our previous results to evaluate each of the terms appearing in Eq. (2.269). From the velocity relations given by Eq. (2.250) and the boundary conditions given by Eq. (2.251), it is seen that

$$\begin{array}{rl} \displaystyle\left( {\frac{{\partial u_r }}{{\partial r}}} \right)_{r = b/2} =& \displaystyle\left[ {\frac{1}{r}\left( { - \frac{1}{r}\frac{{\partial \psi }}{{\partial \theta }} + \frac{{\partial ^2 \psi }}{{\partial r\partial \theta }}} \right)} \right]_{r = b/2} = \left[ {\frac{1}{r}\left( { - u_r - \frac{{\partial u_\theta }}{{\partial \theta }}} \right)} \right]_{r = b/2} \\ {}=& \displaystyle\frac{2}{b}\left( { - u_o \cos \theta e\,^{j\omega t} + u_o \frac{\partial }{{\partial \theta }}\left( {\sin \theta } \right)e\,^{j\omega t} } \right) = 0 \\ \displaystyle\left( {\frac{1}{r}\frac{{\partial u_r }}{{\partial \theta }}} \right)_{r = b/2} =& \displaystyle\left( {\frac{1}{{r^2 }}\frac{{\partial ^2 \psi }}{{\partial \theta ^2 }}} \right)_{r = b/2} = \frac{2}{b}\frac{\partial }{{\partial \theta }}\left( {u_o \cos \theta e\,^{j\omega t} } \right) \\ {}=& \displaystyle\frac{2}{b}\left( { - u_o \sin \theta e\,^{j\omega t} } \right) = \left( {\frac{{u_\theta }}{r}} \right)_{r = b/2} . \\ \end{array}$$

((2.270))

Furthermore, from Eqs. (2.248) to (2.250) and Eq. (2.270),

$$\begin{array}{rl} \displaystyle\left( {\frac{{\partial u_\theta }}{{\partial r}}} \right)_{r = b/2} =&\displaystyle \left( { - \frac{{\partial ^2 \psi }}{{\partial r^2 }}} \right)_{r = b/2} = \left( {\frac{1}{r}\frac{{\partial \psi }}{{\partial r}}} \right)_{r = b/2} + \left( {\frac{1}{{r^2 }}\frac{{\partial ^2 \psi }}{{\partial \theta ^2 }}} \right)_{r = b/2} - \frac{{\rho _f }}{{\mu _f }}\frac{{\partial \psi _2 }}{{\partial t}} \\ {}=&\displaystyle \left( { - \frac{{u_\theta }}{r}} \right)_{r = b/2} + \left( {\frac{{u_\theta }}{r}} \right)_{r = b/2} - \frac{{\rho _f }}{{\mu _f }}\frac{{\partial \psi _2 }}{{\partial t}} \\ {}=&\displaystyle - \frac{{\rho _f }}{{\mu _f }}\frac{{\partial \psi _2 }}{{\partial t}}. \\ \end{array}$$

((2.271))

Using Eqs. (2.270) and (2.271) in Eq. (2.269), we obtain

$$\begin{array}{rl} P_r =& - \left( p \right)_{r = b/2} \\ P_\theta =& - \rho _f \dfrac{{\partial \psi _2 }}{{\partial t}}. \\ \end{array}$$

((2.272))

Then, Eq. (2.268) becomes

$$F = - L\frac{b}{2}\int\limits_0^{2\pi } {\left( p \right)_{r = b/2} \cos \theta d\theta } + L\rho _f \frac{b}{2}\int\limits_0^{2\pi } {\frac{{\partial \psi _2 }}{{\partial t}}\sin \theta d\theta } .$$

((2.273))

We now use integration by parts on the first integral of Eq. (2.273) to obtain

$$\begin{array}{rl} F_1 =&\displaystyle - L\frac{b}{2}\left[ { {\sin \theta } \Big|_0^{2\pi } - \int\limits_0^{2\pi } {\left( {\frac{{dp}}{{d\theta }}} \right)_{r = b/2} \sin \theta d\theta } } \right] \\ {}=&\displaystyle L\frac{b}{2}\int\limits_0^{2\pi } {\left( {\frac{{dp}}{{d\theta }}} \right)_{r = b/2} \sin \theta d\theta } . \\ \end{array}$$

((2.274))

It can be shown that (Stokes 1901, p. 39)

$$\frac{{\partial p}}{{\partial \theta }} = \rho _f \frac{\partial }{{\partial t}}\left( {r\frac{{\partial \psi _1 }}{{\partial r}}} \right).$$

Therefore, Eq. (2.274) becomes

$$F_1 = L\rho _f \frac{b}{2}\int\limits_0^{2\pi } {\left( {\frac{\partial }{{\partial t}}\left( {r\frac{{\partial \psi _1 }}{{\partial r}}} \right)} \right)_{r = b/2} \sin \theta d\theta } $$

((2.275))

and Eq. (2.273) can be written as

$$F = L\rho _f \frac{b}{2}\frac{\partial }{{\partial t}}\int\limits_0^{2\pi } {\left( {r\frac{{\partial \psi _1 }}{{\partial r}} + \psi _2 } \right)_{r = b/2} \sin \theta d\theta } .$$

((2.276))

Substituting Eqs. (2.252), (2.266), and (2.267) into Eq. (2.276), we arrive at

$$\begin{array}{rl} F =&\displaystyle j\omega L\rho _f e\,^{j\omega t} \frac{b}{2}\int\limits_0^{2\pi } {\left( {r\frac{{\partial \Psi _1 }}{{\partial r}} + \Psi _2 } \right)_{r = b/2} \sin ^2 \theta d\theta } \\ {}=&\displaystyle j\omega L\rho _f \pi e\,^{j\omega t} \frac{b}{2}\left( {r\frac{{\partial \Psi _1 }}{{\partial r}} + \Psi _2 } \right)_{r = b/2} \\ {}=&\displaystyle - j\omega u_o L\rho _f \pi e\,^{j\omega t} \frac{{b^2 }}{2}\left( {\frac{b}{{4r}}\left( {\frac{2}{\lambda }\frac{{K_1 \left(\lambda \right)}}{{K_0 \left(\lambda \right)}} + 1} \right) + \frac{{K_1 \left(2\lambda r/b\right)}}{{\lambda K_0 \left(\lambda \right)}}} \right)_{r = b/2} \\{} =&\displaystyle - j\omega u_o e\,^{j\omega t} M_f \Gamma _{cir} \left(\omega \right) \\ \end{array}$$

((2.277))

where

$$\begin{array}{rl} M_f =&\displaystyle \rho _f \pi \frac{{b^2 }}{4}L \\ \Gamma _{cir} \left(\omega \right) =&\displaystyle \left( {1 + \frac{4}{\lambda }\frac{{K_1 \left(\lambda \right)}}{{K_0 \left(\lambda \right)}}} \right) \\ \end{array}$$

((2.278))

The quantity M _f is the mass of the fluid displaced by the cylinder and Γ _cir is a complex quantity called a hydrodynamic function . The force acting on the cylinder is obtained by taking the real part of F, which is

$$F_f = {\textrm{Real}}\left(F\right) = \omega u_o M_f \left[ {{\textrm{Real}}\left(\Gamma _{cir} \right)\sin \left(\omega t\right) + {\textrm{Imag}}\left(\Gamma _{cir} \right)\cos \left(\omega t\right)} \right].$$

((2.279))

If the harmonic displacement of the cylinder in the x-direction is given by $x = X_o \sin \left(\omega t\right)$, then Eq. (2.279) can be written as

$$F_f = - M_f\, {\textrm{Re}} \left(\Gamma _{cir} \right)\ddot x + M_f\, {\textrm{Im}} \left(\Gamma _{cir} \right)\dot x$$

((2.280))

where the over dot indicate the derivate with respect to time and we have recognized that for harmonic oscillations $u_o = \omega X_o$. On the other hand, if the harmonic displacement of the cylinder in the x-direction is given by $x = X_o e^{j\omega t}$, then Eq. (2.277) can be written as

$$F = X_o \omega ^2 e\,^{j\omega t} M_f \Gamma _{cir} \left(\omega \right).$$

((2.281))