Study of the tip mass and interaction force effects on the frequency response and mode shapes of the AFM cantilever

Abstract

The vibrational characteristics of an atomic force microscope (AFM) cantilever beam play a key role in dynamic mode of the atomic force microscope. As the oscillating AFM cantilever tip approaches the sample, the tip–sample interaction force influences the cantilever dynamics. In this paper, we present a detailed theoretical analysis of the frequency response and mode shape behavior of a cantilever beam in the dynamic mode subject to changes in the tip mass and the interaction regime between the AFM cantilever system and the sample. We consider a distributed parameter model for AFM and use Euler–Bernoulli method to derive an expression for AFM characteristics equation contains tip mass and interaction force terms. We study the frequency response of AFM cantilever under variations of interaction force between AFM tip and sample. Also, we investigate the effect of tip mass on the frequency response and also the quality factor and spring constant of each eigenmodes of AFM micro-cantilever. In addition, the mode shape analysis of AFM cantilever under variations of tip mass and interaction force is investigated. This will incorporate the presentation of explicit analytical expressions and numerical analysis. The results show that by considering the tip mass, the resonance frequencies of the cantilever are decreased. Also, the tip mass has a significant effect on the mode shape of the higher eigenmodes of the AFM cantilever. Moreover, tip mass affects the quality factor and spring constant of each modes.

Appendix

The solution of Eq. (15) and its time derivatives can be expressed as:

$$ \begin{array}{*{20}{c}} {W(x) = {{C}_{1}}\sin ax + {{C}_{2}}\cos ax + {{C}_{3}}\sinh ax + {{C}_{4}}\cosh ax} \hfill \\ {W\prime (x) = a\left( {{{C}_{1}}\cos ax - {{C}_{2}}\sin ax + {{C}_{3}}\cosh ax + {{C}_{4}}\sinh ax} \right)} \hfill \\ {W\prime \prime (x) = {{a}^{2}}\left( { - {{C}_{1}}\sin ax - {{C}_{2}}\cos ax + {{C}_{3}}\sinh ax + {{C}_{4}}\cosh ax} \right)} \hfill \\ {W\prime \prime \prime (x) = {{a}^{3}}\left( { - {{C}_{1}}\cos ax + {{C}_{2}}\sin ax + {{C}_{3}}\cosh ax + {{C}_{4}}\sinh ax} \right)} \hfill \\ \end{array} $$

(27)

From boundary conditions, we have:

$$ W(0) = 0 \Rightarrow {C_4} = - {C_2} $$

(28)

$$ W\prime (0) = 0 \Rightarrow {C_3} = - {C_1} $$

(29)

$$ W\prime \prime (L) = 0 \Rightarrow \left[ {\sin aL + \sinh aL} \right]{C_1} + \left[ {\cos aL + \cosh aL} \right]{C_2} = 0 $$

(30)

And finally:

$$ \begin{array}{*{20}{c}} { - {{m}_{{\text{e}}}}\omega _{n}^{2}\left( {D + \left[ {\sin aL - \sinh aL} \right]{{C}_{1}} + \left[ {\cos aL - \cosh aL} \right]{{C}_{2}}} \right)} \hfill \\ { - {{a}^{3}}{\text{EI}}\left( { - \left[ {\cos aL + \cosh aL} \right]{{C}_{1}} + \left[ {\sin aL - \sinh aL} \right]{{C}_{2}}} \right)} \hfill \\ { + {{k}_{{{\text{ts}}}}}\left( {\left[ {\sin aL - \sinh aL} \right]{{C}_{1}} + \left[ {\cos aL - \cosh aL} \right]{{C}_{2}}} \right) = 0} \hfill \\ \end{array} $$

(31)

by defining ${B_1},\;{B_2},\;{B_3},$ and ${B_4}$ as:

$$ \begin{array}{*{20}{c}} {{{B}_{1}} = \sin aL + \sinh aL} \\ {{{B}_{2}} = \cos aL + \cosh aL} \\ {{{B}_{3}} = \left[ {\sin aL - \sinh aL} \right]\left[ {{{k}_{{{\text{ts}}}}} - {{m}_{{\text{e}}}}\omega _{n}^{2}} \right] + {{a}^{3}}{\text{EI}}\left[ {\cos aL + \cosh aL} \right]} \\ {{{B}_{4}} = \left[ {\cos aL - \cosh aL} \right]\left[ {{{k}_{{{\text{ts}}}}} - {{m}_{{\text{e}}}}\omega _{n}^{2}} \right] - {{a}^{3}}{\text{EI}}\left[ {\sin aL - \sinh aL} \right]} \\ \end{array} $$

(32)

Substituting Eq. (32) in (28, 29, 30, and 31) gives:

$$ {\left[ {\begin{array}{*{20}c} {{C_{1} }} \\ {{C_{2} }} \\ \end{array} } \right]} = {\left[ {\begin{array}{*{20}c} {{B_{1} }} {{B_{2} }} \\ {{B_{3} }} {{B_{4} }} \\ \end{array} } \right]}^{{ - 1}} {\left[ {\begin{array}{*{20}c} {0} \\ {E} \\ \end{array} } \right]} $$

(33)

Where:

$$ E = \frac{{{m_{\rm{e}}}{a^4}{\text{EI}}}}{\rho }D $$

(34)

And the characteristics equation is expressed by:

$$ {B_2}{B_3} - {B_1}{B_4} = 0 $$

(35)