Nonlinear coupling of transverse modes of a fixed-fixed microbeam under direct and parametric excitation

Nonlinear Dyn: manuscript No. (will be inserted by the editor) Nonlinear coupling of transverse modes of a fixed-fixed microbeam under direct and parametric excitation Prashant N. Kambali Ashok Kumar Pandey This is the author s copy of accepted manuscript on 6 September 016. The copyright belongs to Nonlinear Dynamics Journal Springer. The final publication is available at Springer via http://dx.doi.org/10.1007/s11071-016-3114-5 Abstract Tuning of linear frequency and nonlinear frequency response of microelectromechanical systems (MEMS) are important in order to obtain high operating bandwidth. Linear frequency tuning can be achieved through various mechanisms such as heating softening due to DC voltage etc. Nonlinear frequency response is influenced by non-linear stiffness quality factor and forcing. In this paper we present the influence of nonlinear coupling in tuning the nonlinear frequency response of two transverse modes of a fixed-fixed microbeam under the influence of direct and parametric forces near and below the coupling regions. To do the analysis we use nonlinear equation governing the motion along in-plane and out-of-plane directions. For a given DC and AC forcing we obtain static and dynamic equations using the Galerkin s method based on first mode approximation under the two different resonant conditions. First we consider one-to-one internal resonance condition in which the linear frequencies of two transverse modes show coupling. Second we consider the case in which the linear frequencies of two transverse modes are uncoupled. To obtain the non-linear frequency response under both the conditions we solve the dynamic equation with the method of multiple scale (MMS). After validating the results obtained using MMS with the numerical simulation of modal equation we discuss the influence of linear and nonlinear coupling on the frequency response of the in-plane and out-of-plane motion of fixed-fixed beam. We also analyzed the influence of quality factor on the frequency response of the beams near the coupling region. We found that the nonlinear response shows single curve near the coupling region with wider width for low value of quality factor and it shows two different curves when the quality fac- Prashant N. Kambali Department of Mechanical and Aerospace Engineering Indian Institute of Technology Hyderabad Kandi 5085 Telangana India E-mail: me1p1004@iith.ac.in Ashok Kumar Pandey Department of Mechanical and Aerospace Engineering Indian Institute of Technology Hyderabad Kandi 5085 Telangana India E-mail: ashok@iith.ac.in

tor is high. Consequently we can effectively tune the quality factor and forcing to obtain different types of coupled response of two modes of a fixed-fixed microbeam. 1 Introduction Microelectromechanical systems (MEMS) have been the subject of intense research in the design of sensitive sensors and actuators. The performance of MEMS based sensors and actuators are mainly dependent on their resonance frequencies. Hence it is important to study the linear and nonlinear frequency tuning of such devices. While the linear frequency tuning can be achieved through various mechanisms such as hardening due to residual stresses softening due to heating and DC voltage and their combined effect etc. the nonlinear frequency response can be tuned due to nonlinear stiffness quality factor and forcing 1 5. In this paper we discuss about the tuning of nonlinear frequency response of a fixed-fixed microbeam under the direct and parametric excitation by controlling the linear and nonlinear stiffness through the coupling of two transverse modes and their quality factors at different excitation force. Many researchers have analyzed the linear and nonlinear frequency response of in-plane or out-of-plane motion of a fixed-fixed or cantilever beam using one degree of freedom model. Younis and Nayfey 6 and Nayfeh et al. 7 studied the linear and nonlinear response of out-of-plane motion of a fixed-fixed beam subjected to direct forcing. Dumitru et al. 8 analyzed the nonlinear behavior of out-of-plane motion of a cantilever beam subjected to the fringing and direct forces. Linzon et al. 10 studied the parametric response of a cantilever beam in out-of-plane direction under the influence of fringing forces from two symmetrically placed side electrodes. Gutschmidt and Gottlieb 11 1 studied the in-plane motion of an array of fixed-fixed beams under the parametric excitation. Lifshitz et al. 13 analyzed the parametric response of an array of fixed-fixed beams along in-plane direction and compare the results with experiments conducted by Buks and Roukes 14. Kambali and Pandey 15 studied nonlinear response of out-ofplane motion of a fixed-fixed beam under the combined effect of direct and fringing forces. While the direct force leads to the nonlinear Duffing response the fringing force induces parametric response. The combined effect of direct and fringing force results in the nonlinear Duffing response with enhanced amplitude as well as frequency width. To analyze the influence of the coupling of two or more modes Nayfeh et al. 16 analyzed the nonlinear response of longitudinal and transverse motion of a taut string subjected to end excitation. Daqaq et al. 17 studied the linear and nonlinear coupled behavior of torsional micromirror when its torsional and transverse modes show :1 internal resonance condition. Isacsson et al. 18 presented numerical and analytical study of the linear and nonlinear coupled behavior of longitudinal and transverse motion of an array of carbon nanotube with fixed-free condition under parametric excitation. Samanta et al. 19 studied nonlinear coupling between various transverse modes of a MoS nanomechanical beam under 1:1 1: and 1:3 internal resonance conditions. Recently Ramini et al. 9 presented experimental studies of primary and parametric resonances of a MEMS arch resonator. Wie et al. 0 investigated the weak and strong coupling in a periodically driven Duffing resonator elastically coupled to a van der Pol oscillator under 1:1 internal resonance condition. Matheny et al. 1 studied

3 intra- and intermodal nonlinear coupling of a doubly clamped piezoelectric beam. Westra et al. presented theoretical and experimental studies of nonlinear intermodal coupling between the flexural vibration modes of a single clamped-clamped beam. Conley et al. 3 analyzed the nonlinear dynamics of fixed-fixed nanowire and found the transition from a planer motion to whirling motion on increasing the excitation amplitude. Mahboob et al. 4 analyzed the nonlinear coupling of nanomechanical resonators by the coupled Vander Pol-Duffing equations. In this paper we model and analyze the nonlinear coupling of two transverse modes of a fixed-fixed beam under the condition of 1:1 internal resonance. To do nonlinear coupled analysis of in-plane and out-of-plane modes near and away from the coupling region we consider the dimensions and properties of a fixed-fixed beam separated by two side electrodes and a bottom electrode as described by Kambali et al.. The electrostatic force along the out-of-plane direction is based on direct forcing between the bottom electrode and the beam and the parametric forcing between the beam and side electrodes. The force along in-plane direction is pure parametric forcing between the beam and the symmetrically placed side electrodes. Under the electrostatic forcing we apply Galerkin s method to governing equations along two directions and obtain the reduced order form of corresponding static and dynamics equations. To obtain the condition of coupling we take appropriate value of DC voltage such that the linear frequencies of in-plane mode ω 1 and out-of-plane mode ω 1 show coupling. To obtain the nonlinear coupled response we solve the modal dynamic equation using the method of multiple scales under the condition of ω 1 ω. After validating the multiple scale solution with numerical results obtained by solving modal dynamic equations we analyze the influence of quality factor on nonlinear frequency response near the coupling region. Governing equations To present the partial differential equations governing the in-plane and out-ofplane motions of a fixed-fixed microbeam we consider a beam of length L width B and thickness H which is separated from the two side electrodes E 1 and E by gaps of g 0 and g 1 respectively and the bottom electrode E g by a gap of d as shown in Fig. 1 (a) and (b). Taking the deflection of the beam along in-plane and out-of-plane as y(x t) and z(x t) respectively as shown in Fig. 1(a) the governing equation of motion along in-plane and out-of-plane directions considering damping residual tension and mid-plane stretching 1 can be written as: EI z ȳ + ρa ȳ + C 1 ȳ N 0 + EA L EIȳ z + ρa z + C 3 z N 0 + EA L L 0 L 0 ( z ) + ȳ d x ȳ n = Qȳ(ȳ z t) (.1) ( z ) + ȳ d x z n = Q z (ȳ z t) (.) where the subscripts prime and dot represent differentiation with respect to x and t respectively N 0 is the initial tension induced in the beam by fabrication processes and heating E is the Young s modulus of the beam EI is the bending

4 Fig. 1 (a) A fixed-fixed beam of width B thickness H separated from the side electrodes E 1 and E by g 0 g 1 and the ground electrode E g by distance d are subjected to direct force Q z and fringing field force Q y; (b) Top view of a fixed-fixed beam of length L. separated from the side electrodes E 1 and E by g 0 g 1 respectively. rigidity I z = HB 3 /1 I y = BH 3 /1 are area moment of inertia about z and y- axes and ρ is the material density. The boundary conditions for the fixed-fixed beam are taken as ȳ(0 t) = ȳ(l t) = 0 z(0 t) = z(l t) = 0 ȳ (0 t) = ȳ (L t) = 0 z (0 t) = z (L t) = 0. (.3) The forcing Q y and Q z are the effective electrostatic forces per unit length along y and z directions for the beam under the direct and fringing field effect as shown in Fig. 1 (a). The expressions for the forcing are given by Qȳ(ȳ z t) = 1 k 1ɛ 0 H (V10 + v(t)) (g 0 ȳ) (V 1 + v(t)) (g 1 + ȳ) (.4) Q z (ȳ z t) = 1 V 1g ɛ 0 4.3 B 3 B (d z) + 0.018 B ( d z ) k 0.00068 (d z) 3 1 ɛ 0 H ( k3 g 0 B 0.156 z + 0.0049B ) (V 10 + v(t)) + (V 1 + v(t)) (.5) where ɛ 0 = 8.85 10 1 F/m is the vacuum permittivity. Here k 1 contributes for the net effect of fringing and direct fields in y-direction k and k 3 represent the strength of the fringing field effects from the bottom electrode and two side electrodes in the z-direction deflection. Further details regarding the extraction of electrostatic force parameters can be found in 5. V ij = V i V j is the voltage difference between the beam and electrodes and v(t) = V ac cos(ωt)..1 Non-dimensionalization To obtain the non-dimensional form of the governing equations we define the ratio of the beam and side electrode gaps as r 0 = (g 0 /g 0 ) r 1 = (g 1 /g 0 ) and use the variables x = x/l y = ȳ/g 0 z = z/d t = t/t where T = ρal 4 /EI z. Finally the non-dimensional nonlinear dynamic equations along the in-plane and out-ofplane directions for fixed-fixed beam under direct and parametric excitation can

5 be written as: y + ÿ + c 1 ẏ N + α 1 Γ (y y) + α Γ (z z)y = β s (V10 + v(t)) (1 y) (V 1 + v(t)) (r 1 + y) z + α 3 z + c 3 ż α 3 N + α 1 Γ (y y) + α Γ (z z)z = (β g + β g (1 z) (.6) β 3g (1 z) 3 ) (V 1g + v(t)) (1 z) (α g + α g z)(v 10 + v(t)) + (V 1 + v(t)). (.7) The corresponding non-dimensional form of the boundary conditions can be written as y(0 t) = y(1 t) = 0 z(0 t) = z(1 t) = 0 y (0 t) = y (1 t) = 0 z (0 t) = z (1 t) = 0. (.8) The various non-dimensionalised parameters as mentioned in above equations are defined as Γ (p(x t) q(x t)) = 1 0 p q x x dx N = N 0L α 1 = 6g 0 EI z B α = 6d B α 3 = β s = 6k 1σ 1 B 3 g0 3 β g = 5.9σ 1 H 3 d 3 β g = 0.109σ 1 B H 3 d β 3g = k 4.08 10 3 σ 1 H 3 B 3 σ 1 = ɛ 0L 4 E α g = 0.094σ 1 g 0 B H d α g = k 39.36 10 1 σ 1 g 0 B 3 H c 1 = C 1L 4 EIz T c 3 = C 3L 4. (.9) EIy T α 3 Since the beam deflection is based on its static component due to DC voltage and dynamic component due to AC voltage the deflection along y and z directions can be written as 1 ( Iz y(x t) = u s (x) + u(x t) z(x t) = w s (x) + w(x t). (.10) where u s (x) and w s (x) are static deflections. By substituting Eqn. (.10) in Eqns. (.6) and (.7) and subsequently setting the time derivatives and dynamic forcing terms equal to zero we get the static equations as: u s N + α 1 Γ (u s u s ) + α Γ (w s w s ) u (V10 ) s = β s (1 u s ) (V 1) (r 1 + u s ) I y ) (.11) w s α 3 N + α 1 Γ (u s u s ) + α Γ (w s w s ) w s = (β g + β g (1 w s ) β 3g (1 w s ) 3 (V 1g ) ) (1 w s ) α g(v 10 ) + (V 1 ) α g (V 10 ) w s + (V 1 ) w s. (.1) Similarly substituting Eqn. (.10) in Eqns. (.6) and (.7) expanding the forcing terms about u n = 0 and w n = 0 upto the first order and subtracting the contribution of the nonlinear static terms given by Eqns. (.11) and (.1) we get the

6 nonlinear dynamic equations for beam as u + u tt + c 1 u t α 1 Γ (u u) + α 1 Γ (u s u) + α Γ (w w) + α Γ (w s w)u s N + α 1 Γ (u s u s ) + α 1 Γ (u u) + α 1 Γ (u s u) + α Γ (w s w s ) + α Γ (w w) +α Γ (w s w)u (V10 ) u = β s (1 u s ) 3 + (V 1) u V 10 v(t) + v(t) (r 1 + u s ) 3 + β s (1 u s ) 1 + u (1 u s ) β s V 1 v(t) + v(t) (r 1 + u s ) 1 u (r 1 + u s ) (.13) w + α 3 w tt + c 3 w t α 3 α 1 Γ (u u) + α 1 Γ (u s u) + α Γ (w w) + α Γ (w s w) w s α 3 N + α 1 Γ (u s u s ) + α 1 Γ (u u) + α 1 Γ (u s u) + α Γ (w s w s ) + α Γ (w w) +α Γ (w s w)w = (β g + β 3g (1 w s ) 3 )w (V 1g) 1 + V 1g v(t) + v(t) (1 w s ) (1 w s ) 3 + β g w β g (V 1g v(t) + v(t) ) β 3g (V 1g v(t) + v(t) )(1 w s (1 w s ) w) α g w(v 10 ) + (V 1 ) (V 10 v(t) + v(t) )(α g α g w s α g w) (V 1 v(t) + v(t) )(α g α g w s α g w) (.14). Reduced order model To obtain the static and dynamic equations to perform coupled analysis under 1:1 internal resonance condition near the coupling region which is much below the pull-in voltage 3 we use Galerkin method based on first mode approximation of the in-plane and out-of-plane displacements with negligible error 6. Assuming the static and dynamic displacements subjected to the first transverse mode φ(x) the displacement along in-plane and out-of-plane directions can be written as 3: u s (x) = q 1 (y z)φ(x) w s (x) = q (y z)φ(x) (.15) u(x t) = P 1 (t)φ(x) w(x t) = P (t)φ(x) (.16) where q 1 and q are static deflections and P 1 (t) and P (t) are non-dimensional modal variables. φ(x) is ( undamped exact mode ) shape which is taken as φ(x) = cosh(ζx) cos(ζx) ν sinh(ζx) sin(ζx) where for the first mode ζ = 4.73 and ν = 0.985 such that 1 0 (φ 1(x)) dx = 1. After premultiplying the denominator terms on either side of the Eqns (.11) (.1) substituting the assumed solution given by Eqn. (.15) and applying Galerkin s method we obtain the nonlinear static equations in both the directions which are given in Appendix A. Similarly by premultiplying the denominator terms on either side of the Eqns. (.13) and (.14) substituting the assumed static and dynamic solutions given by Eqns. (.15) and (.16) and applying Galerkin s method we obtain the nonlinear

7 modal dynamic equations in both the directions as: P 1tt (t) + λ 1P 1 (t) + t 1 P 1 (t) 3 + t P 1 (t) + t 3 P (t) + t 4 P (t) + t 5 (V 10 v(t) +v(t) ) + t 6 (V 1 v(t) + v(t) ) P 1 (t) + t 11 P 1t (t) + t 7 P (t) + t 8 P (t) +t 9 (V 10 v(t) + v(t) ) + t 10 (V 1 v(t) + v(t) ) = 0 (.17) P tt (t) + λ P (t) + s 1 P (t) 3 + s P (t) + s 3 P (t) + s 4 P (t) + s 5 (V 1g v(t) +v(t) ) + s 6 (V 10 v(t) + V 1 v(t) + v(t) ) P 1 (t) + s 11 P t (t) + s 7 P 1 (t) +s 8 P 1 (t) + s 9 (V 1g v(t) + v(t) ) + s 10 (V 10 v(t) + V 1 v(t) + v(t) ) = 0 (.18) where all coefficients of each term in above Eqns. (.17) and (.18) are given in Appendix A. Neglecting the damping term nonlinear terms and the dynamic forcing terms from above Eqns. (.17) and (.18) we get linear modal dynamic equations as: P 1tt (t) + λ 1P 1 (t) + t 8 P (t) = 0 (.19) P tt (t) + λ P (t) + s 8 P 1 (t) = 0. (.0) To obtain the linear frequency from the above equation we assume the solution of Eqns.(.19) and (.0) as P 1 (t) = βe iωt P (t) = γe iωt. Substituting the assumed solutions in the modal Eqns.(.19) and (.0) we get (λ 1 ω )β + t 8 γ = 0 (λ ω )γ + s 8 β = 0 For non-trivial solution the determinant of these system of equations should be zero. After solving the resulting equation we get two values of ω corresponding to two directions as 1 ω 1 = (λ (λ 1 + λ ) ± 1 λ ) + 4t 8 s 8 where λ 1 λ t 8 and s 8 are given in Appendix A. (.1) 3 Method of multiple scale In this section we apply the method of multiple scales (MMS) in solving Eqns. (.17) and (.18) assuming the modal displacements as functions multiple time scales T 0 = t T 1 = ɛt and T = ɛ t as P 1 (t) = ɛx 11 (T 0 T 1 T ) + ɛ x 1 (T 0 T 1 T ) + ɛ 3 x 13 (T 0 T 1 T ) + O(ɛ 4 ) P (t) = ɛx 1 (T 0 T 1 T ) + ɛ x (T 0 T 1 T ) + ɛ 3 x 3 (T 0 T 1 T ) + O(ɛ 4 ) (3.1)

8 where ɛ is a dimensionless small positive number. Subsequently the derivative terms with respect to t can be defined in terms of new time scales as d dt = dt 0 T 0 dt + dt 1 T 1 dt + ( d dt d dt = dt = (D 0 + ɛd 1 + ɛ D ) T dt ) = (D 0 + ɛd 1 + ɛ D ) = (D0 + ɛd 0 D 1 + ɛ D1 +ɛ D 0 D ) + H.O.T (3.) Rescaling the damping and forcing terms with different powers of ɛ as t 11 = ɛt 11 s 11 = ɛs 11 v(t) = ɛ V ac cos(ωt) (3.3) substituting the assumed solution from Eqns. (3.1) to (3.3) in Eqns (.17) and (.18) and by comparing different powers of ɛ upto third order we get the following three sets of equations as O(ɛ 1 ) D 0x 11 + λ 1x 11 + t 8 x 1 = 0 D 0x 1 + λ x 1 + s 8 x 11 = 0 (3.4) O(ɛ ) D0x 1 + λ 1x 1 + t 8 x = D 0 D 1 x 11 t 11 D 0 x 11 t x 11 t 4 x 11 x 1 t 7 x 1 t 9 η 11 cos(ω ac t) t 10 η 1 cos(ω ac t) D0x + λ x + s 8 x 1 = D 0 D 1 x 1 s 11 D 0 x 1 s x 1 s 4 x 11 x 1 s 7 x 11 s 9 η 1 cos(ω ac t) s 10 (η 11 + η 1 ) cos(ω ac t) (3.5) O(ɛ 3 ) D0x 13 + λ 1x 13 + t 8 x 3 = t 11 (D 0 x 1 + D 1 x 11 ) (D 0 D 1 x 1 +D1x 11 + D 0 D x 11 ) t 1 x 3 11 t x 11 x 1 t 3 x 11 x 1 t 4 (x 11 x + x 1 x 1 ) t 5 η 11 cos(ω ac t)x 11 t 6 η 1 cos(ω ac t)x 11 t 7 x 1 x D0x 3 + λ x 3 + s 8 x 13 = s 11 (D 0 x + D 1 x 1 ) (D 0 D 1 x + D1x 1 +D 0 D x 1 ) s 1 x 3 1 s x 1 x s 3 x 1 x 11 s 4 (x 11 x + x 1 x 1 ) s 5 η 1 cos(ω ac t)x 1 s 6 (η 11 + η 1 ) cos(ω ac t)x 1 s 7 x 11 x 1 (3.6) where η 11 = V 10 V ac η 1 = V 1 V ac and η 1 = V 1g V ac. 3.1 Solution of 1 st order equation Solutions x 11 and x 1 for the two homogeneous second order coupled equations given by Eqn. (3.4) are can be written as x 11 = A 1 (T 1 T )e iω 1T 0 + A (T 1 T )e iω T 0 + Ā1(T 1 T )e iω 1T 0 + Ā(T 1 T )e iω T 0 x 1 = k 1 A 1 (T 1 T )e iω 1T 0 + k A (T 1 T )e iω T 0 + k 1 Ā 1 (T 1 T )e iω 1T 0 +k Ā (T 1 T )e iω T 0 (3.7)

9 where ω 1 and ω are the coupled natural frequencies of the system in two orthogonal directions obtained from linear analysis. Substituting the assumed form of the solution from Eqn.(3.7) into Eqn.(3.4) we get (λ 1 ω 1) + t 8 k 1 A 1 e iω 1T 0 + (λ 1 ω ) + t 8 k A e iω T 0 = 0 (λ ω 1)k 1 + s 8 A 1 e iω 1T 0 + (λ ω )k + s 8 A e iω T 0 = 0. (3.8) Equating the coefficients of different terms on both sides of Eqn.(3.8) we get (λ 1 ω 1) + t 8 k 1 = 0 (λ 1 ω ) + t 8 k = 0 (3.9) (λ ω 1)k 1 + s 8 = 0 (λ ω )k + s 8 = 0. (3.10) On solving Eqns.(3.9) and (3.10) for k 1 and k we get the solvability condition as ( ω k n = n λ ) ( ) 1 s8 = t 8 ωn λ (3.11) where n = 1 and. 3. Solution of nd order equation To obtain the solution of Eqn. (3.5) we substitute Eqn. (3.7) into it and eliminate the secular terms. Taking the detuning parameters σ 1 and σ as Ω = ω 1 + ɛσ 1 ω = ω 1 + ɛσ. (3.1) and substituting x 11 and x 1 from Eqn. (3.7) into Eqn. (3.5) we get D 0x 1 + λ 1x 1 + t 8 x = i(ω 1 D 1 A 1 e iω 1T 0 + ω D 1 A e iω T 0 ) it 11 (A 1 ω 1 e iω 1T 0 +A ω e iω T 0 ) t (A 1e iω 1T 0 + A 1 Ā 1 + A e iω T 0 + A Ā + A 1 Ā e i(ω 1 ω )T 0 +A 1 A e i(ω 1+ω )T 0 ) t 4 (A 1k 1 e iω 1T 0 + A 1 Ā 1 k 1 + A k e iω T 0 + A Ā k + A 1 Ā (k 1 + k )e i(ω 1 ω )T 0 + A 1 A (k 1 + k )e i(ω 1+ω )T 0 ) t 7 (k 1A 1e iω 1T 0 + k 1A 1 Ā 1 +k A e iω T 0 + k A Ā + k 1 k A 1 Ā e i(ω 1 ω )T 0 + k 1 k A 1 A e i(ω 1+ω )T 0 ) + t 9 η 11e iω act 0 t 10 η 1e iω act 0 + cc (3.13) D 0x + λ x + s 8 x 1 = i(ω 1 k 1 D 1 A 1 e iω 1T 0 + k ω D 1 A e iω T 0 ) is 11 (k 1 A 1 ω 1 e iω 1T 0 + k A ω e iω T 0 ) s (k 1A 1e iω 1T 0 + k 1A 1 Ā 1 + k A e iω T 0 + k A Ā +k 1 k A 1 Ā e i(ω 1 ω )T 0 + k 1 k A 1 A e i(ω 1+ω )T 0 ) s 4 (A 1k 1 e iω 1T 0 + A 1 Ā 1 k 1 +A k e iω T 0 + A Ā k + A 1 Ā (k 1 + k )e i(ω 1 ω )T 0 + A 1 A (k 1 + k )e i(ω 1+ω )T 0 ) s 7 (A 1e iω 1T 0 + A 1 Ā 1 + A e iω T 0 + A Ā + A 1 Ā e i(ω 1 ω )T 0 + A 1 A e i(ω 1+ω )T 0 ) + s 9 η 1e iω act 0 s 10 (η 11 + η 1 )e iω act 0 + cc. (3.14) Assuming the solution of homogeneous form of Eqn.(3.5) as x 1 = P 11 e iω 1T 0 + P 1 e iω T 0 + cc x = P 1 e iω 1T 0 + P e iω T 0 + cc (3.15)

10 and substituting it in Eqns. (3.13) and (3.14) and eliminating the secular terms we get (λ 1 ω 1)P 11 e iω 1T 0 + (λ 1 ω )P 1 e iω T 0 + t 8 (P 1 e iω 1T 0 + P e iω T 0 ) = R 11 e iω 1T 0 + R 1 e iω T 0 (3.16) (λ ω 1)P 1 e iω 1T 0 + (λ ω )P e iω T 0 + s 8 (P 11 e iω 1T 0 + P 1 e iω T 0 ) The above equations can also be written in the matrix form as (λ 1 ωn) t 8 P1n s 8 (λ ωn) = P n = R 1 e iω 1T 0 + R e iω T 0. (3.17) R1n where n = 1 and R 1n and R n are the coefficients of e iω 1T 0 and e iω T 0 appearing in Eqns. (3.13) and (3.14) as R n R 11 = iω 1 D 1 A 1 iω D 1 A e iσ T 1 it 11 A 1 ω 1 it 11 A ω e iσ T 1 ( ) t9 η 11 + t 10 η 1 e iσ 1T 1 R 1 = iω 1 D 1 A 1 e iσ T 1 iω D 1 A it 11 A 1 ω 1 e iσ T 1 it 11 A ω ( ) t9 η 11 + t 10 η 1 e i(σ 1 σ )T 1 R 1 = iω 1 k 1 D 1 A 1 ik ω D 1 A e iσ T 1 is 11 k 1 A 1 ω 1 is 11 k A ω e iσ T 1 ( ) s9 η 1 + s 10 (η 11 + η 1 ) e iσ 1T 1 R = iω 1 k 1 D 1 A 1 e iσ T 1 iω k D 1 A is 11 k 1 A 1 ω 1 e iσ T 1 is 11 k A ω ( ) s9 η 1 + s 10 (η 11 + η 1 ) e i(σ 1 σ )T 1. (3.18) Solving Eqns. (3.16) and (3.17) and using Eqn. (3.11) we get the solvability conditions in terms of R 1n and R n as R 1n = t 8 (λ ω n) R n = ( t8 s 8 ) k n R n = k n R n (3.19) where kn = ( t 8 s 8 )k n. For n = 1 and k 1 = ( t 8 s 8 )k 1 the solvability condition R 11 + k 1 R 1 = 0 reduces to iω 1 (1 + k 1 k1 )D 1 A 1 iω (1 + k k1 )D 1 A e iσ T 1 = iω 1 A 1 (t 11 + s 11 k 1 k1 ) + iω A (t 11 + s 11 k k1 )e iσ T 1 ( t9 η 11 + t 10 η 1 + + k 1 s 9 η 1 + k ) 1 s 10 (η 11 + η 1 ) e iσ 1T 1. (3.0)

11 Similarly for n = and k = ( t 8 s 8 )k the solvability condition R 1 + k R = 0 can be written as iω 1 (1 + k 1 k )D 1 A 1 e iσ T 1 iω (1 + k k )D 1 A = iω A (t 11 + s 11 k k ) + iω 1 A 1 (t 11 + s 11 k 1 k )e iσ T 1 ( t9 η 11 + t 10 η 1 + + k s 9 η 1 + k ) s 10 (η 11 + η 1 ) e i(σ 1 σ )T 1. (3.1) Solving Eqns. (3.0) and (3.1) simultaneously we find the expressions for D 1 A 1 and D 1 A as D 1 A 1 = B 1A 1 + B A e iσ T 1 + ib 3 e iσ 1T 1 B 4 D 1 A = B 5A + B 6 A 1 e iσ T 1 + ib 7 e i(σ 1 σ )T 1 B 8 (3.) where B 1 B... B 8 are defined in Appendix B. Now substituting D 1 A 1 and D 1 A from Eqn.(3.) into the second order Eqns. (3.13) and (3.14) we obtain the complete solution for second order equations as x 1 = A 3 (T 1 T )e iω 1T 0 + A 4 (T 1 T )e iω T 0 + c 11 A 1e iω 1T 0 + c 1 A 1 Ā 1 + c 13 A e iω T 0 + c 14 A Ā + c 15 A 1 Ā e i(ω 1 ω )T 0 + c 16 A 1 A e i(ω 1+ω )T 0 + cc x = k 1 A 3 (T 1 T )e iω 1T 0 + k A 4 (T 1 T )e iω T 0 + c 1 A 1e iω 1T 0 + c A 1 Ā 1 + c 3 A e iω T 0 + c 4 A Ā + c 5 A 1 Ā e i(ω 1 ω )T 0 + c 6 A 1 A e i(ω 1+ω )T 0 + cc (3.3) On substituting Eqn.(3.3) in Eqns. (3.13) and (3.14) and comparing the coefficients of same terms on both side of the resulting equations we obtain the following equations in terms of coefficients c ij. Coefficients of A 1e iω 1T 0 : Coefficients of A 1 Ā 1 : Coefficients of A e iω T 0 : Coefficients of A Ā : 4c 11 ω 1 + c 11 λ 1 + t 8 c 1 = t t 4 k 1 t 7 k 1 4c 1 ω 1 + c 1 λ + s 8 c 11 = s 7 s 4 k 1 s k 1 (3.4) λ 1c 1 + t 8 c = t t 4 k 1 t 7 k 1 λ c + s 8 c 1 = s 7 s 4 k 1 s k 1 (3.5) 4c 13 ω + c 13 λ 1 + t 8 c 3 = t t 4 k t 7 k 4c 3 ω + c 3 λ + s 8 c 13 = s 7 s 4 k s k (3.6) λ 1c 14 + t 8 c 4 = t t 4 k t 7 k λ c 4 + s 8 c 14 = s 7 s 4 k s k (3.7)

1 Coefficients of A 1 Ā e i(ω 1 ω )T 0 : c 15 (ω 1 ω ) + λ 1c 15 + t 8 c 5 = t t 4 (k 1 + k ) t 7 k 1 k c 5 (ω 1 ω ) + λ c 5 + s 8 c 15 = s 7 s 4 (k 1 + k ) s k 1 k (3.8) Coefficients of A 1 A e i(ω 1+ω )T 0 : c 16 (ω 1 + ω ) + λ 1c 16 + t 8 c 6 = t t 4 (k 1 + k ) t 7 k 1 k c 6 (ω 1 + ω ) + λ c 6 + s 8 c 16 = s 7 s 4 (k 1 + k ) s k 1 k (3.9) Solving above equations simultaneously the coefficients c ij are obtained which are given in Appendix B. 3.3 Solution of 3 rd order equation Substituting Eqns. (3.7) and (3.3) into Eqn.(3.6) using Ω = ω 1 + ɛσ 1 ω = ω 1 + ɛσ and separating the secular terms similar to previous section we obtain the following equations: R 11 = iω 1 (D 1 A 3 + D A 1 ) D 1A 1 t 11 D 1 A 1 iω (D 1 A 4 + D A ) + D 1A +t 11 D 1 A e iσ T 1 it 11 ω 1 A 3 it 11 ω A 4 e iσ T 1 g 11 A 1Āe iσ T 1 g 1 A Āe iσ T 1 g 13 A A 1 Ā 1 e iσ T 1 g 14 A Ā1e iσ T 1 g 15 A 1Ā1 g 16 A 1 A Ā (3.30) R 1 = iω (D 1 A 4 + D A ) D A t 11 D 1 A iω 1 (D 1 A 3 + D A 1 ) + D 1A 1 t 11 D 1 A 1 e iσ T 1 it 11 ω A 4 it 11 ω 1 A 3 e iσ T 1 f 11 A Ā1e iσ T 1 f 1 A 1Ā1 e iσ T 1 f 13 A 1 A Ā e iσ T 1 f 14 A 1Āe iσ T 1 f 15 A Ā f 16 A A 1 Ā 1 (3.31) R 1 = iω 1 k 1 (D 1 A 3 + D A 1 ) k 1 D 1A 1 s 11 k 1 D 1 A 1 ik ω (D 1 A 4 + D A ) +k D 1A + k s 11 D 1 A e iσ T 1 is 11 k 1 ω 1 A 3 is 11 k ω A 4 e iσ T 1 g 1 A 1Ā1e iσ T 1 (g A Ā + g 3 A A 1 Ā 1 )e iσ T 1 g 4 A Ā1e iσ T 1 g 5 A 1Ā1 g 6 A 1 A Ā (3.3) R = ik ω (D 1 A 4 + D A ) k D 1A s 11 k D 1 A ik 1 ω 1 (D 1 A 3 + D A 1 ) +k 1 D 1A 1 + s 11 k 1 D 1 A 1 e iσ T 1 ik s 11 ω A 4 ik 1 s 11 ω 1 A 3 e iσ T 1 f 1 A Ā1 e iσ T 1 f A 1Ā1e iσ T 1 f 3 A 1 A Ā e iσ T 1 f 4 A 1Āe iσ T 1 f 5 A Ā f 6 A A 1 Ā 1 (3.33) where g n1...g n6 and f n1...f n6 for n = 1 and are given in Appendix B. Substituting the above equations in solvability condition given by Eqn.(3.19) we get the following two conditions iω 1 B 11 (D 1 A 3 + D A 1 ) B 11 D 1A 1 B 13 D 1 A 1 iω B 1 (D 1 A 4 + D A ) +B 1 D 1A + B 14 D 1 A e iσ T 1 = iω 1 B 13 A 3 + iω B 14 A 4 e iσ T 1 + ḡ 1 Ā A 1e iσ T 1 +ḡ Ā A e iσ T 1 + ḡ 3 A 1 A Ā 1 e iσ T 1 + ḡ 4 A Ā1e iσ T 1 + ḡ 5 A 1Ā1 + ḡ 6 A 1 Ā A (3.34)

13 iω G 1 (D 1 A 4 + D A ) G 1 D 1A G 13 D 1 A iω 1 G 11 (D 1 A 3 + D A 1 ) +G 11 D 1A 1 + G 14 D 1 A 1 e iσ T 1 = iω G 13 A 4 + iω 1 G 14 A 3 e iσ T 1 + f 1 Ā 1 A e iσ T 1 + f Ā 1 A 1e iσ T 1 + f 3 A 1 A Ā e iσ T 1 + f 4 A 1Āe iσ T 1 + f 5 A Ā + f 6 A 1 Ā 1 A (3.35) where g n = g 1n + k 1 g n fn = f 1n + k f n and the coefficients B 11 B 1 B 13 B 14 G 11 G 1 G 13 G 14 f 1n and g 1n for n = 1... 6 are mentioned in Appendix B. Now by following the procedure as mentioned in 717 we find A 3 and A 4 so as to eliminate D1A 1 and D1A from Eqns. (3.34) and (3.35). Such conditions lead to the following equations D 1 iω 1 A 3 + D 1 A 1 = 0 iω 1 A 3 + D 1 A 1 = h 11 (T ) D 1 iω A 4 + D 1 A = 0 iω A 4 + D 1 A = h 1 (T ); (3.36) However it is evident from Eqn.(3.) that D 1 A 1 and D 1 A are implicit functions of slow time scale T. Thus we take h 11 (T ) = h 1 (T ) =0. Now using Eqns.(3.) and (3.36) the expressions for A 3 and A 4 can be written as A 3 = ib 1 ω 1 B 4 A 1 + ib ω 1 B 4 A e iσ T 1 B 3 ω 1 B 4 e iσ 1T 1 A 4 = ib 5 ω B 8 A + ib 6 ω B 8 A 1 e iσ T 1 B 7 ω B 8 e i(σ 1 σ )T 1. (3.37) Using Eqn. (3.37) in Eqns. (3.34) and (3.35) we can find the values for D A 1 and D A by solving the following equations iω 1 B 11 D A 1 iω B 1 D A e iσ T 1 = B 13 D 1 A 1 + B 14 D 1 A e iσ T 1 + iω 1 B 13 A 3 +iω B 14 A 4 e iσ T 1 + ḡ 1 Ā A 1e iσ T 1 + ḡ Ā A e iσ T 1 + ḡ 3 A 1 A Ā 1 e iσ T 1 +ḡ 4 A Ā1e iσ T 1 + ḡ 5 A 1Ā1 + ḡ 6 A 1 Ā A (3.38) iω 1 G 11 D A 1 e iσ T 1 iω G 1 D A = G 13 D 1 A + G 14 D 1 A 1 e iσ T 1 +iω G 13 A 4 + iω 1 G 14 A 3 e iσ T 1 + f 1 Ā 1 A e iσ T 1 + f Ā 1 A 1e iσ T 1 + f 3 A 1 A Ā e iσ T 1 + f 4 A 1Āe iσ T 1 + f 5 A Ā + f 6 A 1 Ā 1 A. (3.39) The expressions for D A 1 and D A can be obtained as i D A 1 = (B 14 G 1 B 1 G 13 )D 1 A e iσ T 1 + (B 13 G 1 ω 1 (B 11 G 1 B 1 G 11 ) B 1 G 14 )D 1 A 1 + iω 1 (B 13 G 1 B 1 G 14 )A 3 + iω (B 14 G 1 B 1 G 13 )A 4 e iσ T 1 D A = +G 1 Ḡ B 1 F e iσ T 1 i (B 14 G 11 B 11 G 13 )D 1 A + (B 13 G 11 ω (B 1 G 11 B 11 G 1 ) B 11 G 14 )D 1 A 1 e iσ T 1 + iω (B 14 G 11 B 11 G 13 )A 4 + iω 1 (B 13 G 11 B 11 G 14 )A 3 e iσ T 1 + G 11 Ḡe iσ T 1 B 11 F (3.40)

14 where F and Ḡ are given in the Appendix B. To get the final solution we apply the method of reconstitution 7 17 and write modulation in the following form da 1 = ɛd 1 A 1 + ɛ D A 1 +... dt da = ɛd 1 A + ɛ D A +... (3.41) dt Substituting the values of D 1 A 1 and D 1 A from Eqn. (3.) D A 1 and D A from Eqn. (3.40) and A 3 and A 4 from Eqn.(3.37) into Eqn.(3.41) and setting ɛ = 1 such that T 0 = T 1 = T = t we get the reconstituted modulation equations as A 1 = (B 13 G 1 G 14 B 1 ) 1 + i 4ω 1 (B 11 G 1 G 11 B 1 ) (B14 G 1 G 13 B 1 ) + i 4ω 1 (B 11 G 1 G 11 B 1 ) (G1 Ḡ B 1 F e iσ t ) + i ω 1 (B 11 G 1 G 11 B 1 ) A (B13 G 11 G 14 B 11 ) = i 4ω (B 1 G 11 G 1 B 11 ) e iσ t (B 14 G 11 G 13 B 11 ) + 1 + i 4ω (B 1 G 11 G 1 B 11 ) (G11 Ḡe iσt B 11 F ) + i ω (B 1 G 11 G 1 B 11 ) B1 A 1 + B A e iσt + i B 3 B 4 B 4 B5 e iσ 1t B 4 B 8 A + B 6 B 8 A 1 e iσ t + i B 7 B 8 e i(σ 1 σ )t (3.4) e iσ t B1 B 4 A 1 + B B 4 A e iσ t + i B 3 B 4 e iσ 1t B5 A + B 6 A 1 e iσt + i B 7 e i(σ 1 σ )t B 8 B 8 B 8. (3.43) To express the modulation equations in polar form we rewrite A 1 and A as Here a n and β n are real functions of time t hence A n = 1 a ne iβ n n = 1. (3.44) A 1 = 1 a 1e iβ 1 A 1 = 1 ( a 1e iβ 1 + ia 1 β 1 e iβ 1 ) A 1 and A can be written as A = 1 a e iβ A = 1 ( a e iβ + ia β e iβ ). (3.45) Substituting Eqn.(3.45) into Eqns.(3.4) and (3.43) we get 1 ( a 1e iβ 1 + ia 1 β 1 e iβ 1 ) = ( ) 1 + ih 11 h a 1 e iβ 1 + h 33 a e i(β +σ t) + ih 44 e iσ 1t +ih 55 e iσ t h 66 a e iβ + h 77 a 1 e i(β 1 σ t) + ih 88 e i(σ 1 σ )t + ih 99 ḡ 1 a 1a e i(β 1 β σ t) + ḡ a 3 e i(β +σ t) + ḡ 3 a 1a e i(β +σ t) + ḡ 4 a 1 a e i(β β 1 +σ t) +ḡ 5 a 3 1e iβ 1 + ḡ 6 a 1 a e iβ 1 ih 1010 f 1 a 1 a e i(σ t β 1 +β ) + f a 3 1e iβ 1 + f 3 a 1 a e iβ 1 + f 4 a 1a e i(β 1 β σ t) + f 5 a 3 e i(β +σ t) + f 6 a 1a e i(β +σ t) (3.46)

1 ( a e iβ + ia β e iβ ) = ( ) 1 + il 11 l a e iβ + l 33 a 1 e i(β 1 σ t) + il 44 e i(σ 1 σ )t +il 55 e iσ t l 66 a 1 e iβ 1 + l 77 a e i(β +σ t) + il 88 e iσ 1t + il 99 ḡ 1 a 1a e i(β 1 β σ t) 15 +ḡ a 3 e iβ + ḡ 3 a 1a e iβ + ḡ 4 a 1 a e i(β β 1 +σ t) + ḡ 5 a 3 1e i(β 1 σ t) + ḡ 6 a 1 a il 1010 f 1 a 1 a e i(σ t β 1 +β ) + f a 3 1e i(β 1 σ t) + f 3 a 1 a e i(β 1 σ t) e i(β 1 σ t) + f 4 a 1a e i(β 1 β σ t) + f 5 a 3 e iβ + f 6 a 1a e iβ (3.47) where h 11 h... h 1010 and l 11 l... l 1010 are given in Appendix B. Finally we convert the above non-autonomous equations into autonomous forms by defining two new variables and their corresponding time derivative terms as θ 1 = (σ 1 t β 1 ) θ = (σ 1 σ )t β θ 1 = (σ 1 β 1 ) θ = (σ 1 σ ) β (3.48) Substiting Eqn.(3.48) into Eqns.(3.46) and (3.47) and separating the real and imaginary parts we get the following form of modulation equations a 1 = h a 1 + h 3 a cos(θ 1 θ ) h 4 sin(θ 1 ) + h 1 h 3 a sin(θ 1 θ ) h 4 cos(θ 1 ) + h 5 h 6 a sin(θ 1 θ ) h 8 cos(θ 1 ) + h 9 ḡ 1 a 1a sin(θ 1 θ ) ḡ a 3 sin(θ 1 θ ) ḡ 3 a 1a sin(θ 1 θ ) ḡ 4 a a 1 sin (θ 1 θ ) h 10 f 1 a a 1 sin (θ 1 θ ) + f 4 a 1a sin(θ 1 θ ) f 5 a 3 sin(θ 1 θ ) f 6 a 1a sin (θ 1 θ ) (3.49) a 1 θ 1 = a 1 σ 1 h 3 a sin(θ 1 θ ) h 4 cos(θ 1 ) h 1 h a 1 + h 3 a cos(θ 1 θ ) h 4 sin(θ 1 ) h 5 h 6 a cos(θ 1 θ ) + h 7 a 1 h 8 sin(θ 1 ) h 9 ḡ 1 a 1a cos(θ 1 θ ) + ḡ a 3 cos(θ 1 θ ) + ḡ 3 a 1a cos(θ 1 θ ) + ḡ 4 a a 1 cos (θ 1 θ ) + ḡ 5 a 3 1 + ḡ 6 a 1 a + h 10 f 1 a a 1 cos (θ 1 θ ) + f a 3 1 + f 3 a 1 a + f 4 a 1a cos(θ 1 θ ) + f 5 a 3 cos(θ 1 θ ) + f 6 a 1a cos (θ 1 θ ) (3.50) a = l a + l 3 a 1 cos(θ 1 θ ) l 4 sin(θ ) + l 1 l 3 a 1 sin(θ 1 θ ) l 4 cos(θ ) + l 5 l 6 a 1 sin(θ 1 θ ) l 8 cos(θ ) + l 9 ḡ 1 a 1a sin (θ 1 θ ) ḡ 4 a a 1 sin(θ 1 θ ) + ḡ 5 a 3 1a sin(θ 1 θ ) + ḡ 6 a a 1 sin (θ 1 θ ) l 10 f 1 a a 1 sin (θ 1 θ ) + f a 3 1 sin(θ 1 θ ) + f 3 a 1 a sin(θ 1 θ ) + f 4 a 1a sin (θ 1 θ ) (3.51) a θ = a (σ 1 σ ) + l 3 a 1 sin(θ 1 θ ) l 4 cos(θ ) l 1 l a + l 3 a 1 cos(θ 1 θ ) l 4 sin(θ ) l 5 l 6 a 1 cos(θ 1 θ ) + l 7 a l 8 sin(θ ) l 9 ḡ 1 a 1a cos (θ 1 θ ) + ḡ a 3 + ḡ 3 a a 1 + ḡ 4 a a 1 cos(θ 1 θ ) + ḡ 5 a 3 1 cos(θ 1 θ ) + ḡ 6 a a 1 cos(θ 1 θ ) + l 10 f 1 a a 1 cos (θ 1 θ ) + f a 3 1 cos(θ 1 θ ) + f 3 a 1 a cos(θ 1 θ ) + f 4 a 1a cos (θ 1 θ ) + f 5 a 3 + f 6 a a 1 (3.5)

16 where h 1 h... h 10 and l 1 l...l 10 are given in Appendix B. To obtain the equilibrium solution we set time derivatives terms to zero in Eqns. (3.49) (3.5) and solve the resulting equations. Finally the response of the beam upto second term can be written using Eqns. (3.1) (3.7) (3.3) and (3.44) and ɛ = 1 as ( P 1 (t) = x 11 + x 1 = 1 + i Λ 11 ) ( a 1 cos(ω 1 t + β 1 ) + 1 + i Λ 1 ) a cos(ω t + β ) +Λ 13 cos(ω 1 + σ 1 )t + 1 c 11a 1 cos (ω 1 t + β 1 ) + 1 c 1a 1 + 1 c 13a cos (ω t + β ) + 1 c 14a + 1 c 15a 1 a cos (ω 1 ω )t + β 1 β + 1 c 16a 1 a cos (ω 1 + ω )t + β 1 + β (3.53) ( P (t) = x 1 + x = k 1 1 + i Λ ( 11 )a 1 cos(ω 1 t + β 1 ) + k 1 + i Λ 1 ) a cos(ω t +β ) + Λ 14 cos(ω 1 + σ 1 )t + 1 c 1a 1 cos (ω 1 t + β 1 ) + 1 c a 1 + 1 c 3a cos (ω t + β ) + 1 c 4a + 1 c 5a 1 a cos (ω 1 ω )t + β 1 β + 1 c 6a 1 a cos (ω 1 + ω )t + β 1 + β (3.54) where the terms Λ 11 Λ 1 Λ 13 and Λ 14 are defined as B1 Λ 11 = + B 6 B Λ 1 = + B 5 ω 1 B 4 ω B 8 ω 1 B 4 ω B 8 B3 Λ 13 = + B 7 Λ 14 = ω 1 B 4 ω B 8 k 1 B 3 ω 1 B 4 + k B 7 ω B 8. (3.55) 4 Results and discussion In this section we first study the linear frequency variation of in-plane and outof-plane modes of a microbeam to locate the coupling region. Subsequently we validate the modulation equations developed by the method of multiple scales with numerical solution obtained by solving the modal dynamic equations. Finally we use the method of multiple scale to study coupled nonlinear response near and away from the coupling region. Additionally we also analyze the influence of quality factor on the non-linear frequency response near the coupled region. To do the study we consider the dimensions material properties and electrostatic force coefficients in a fixed-fixed microbeam as mentioned in and are given in Table 1. 4.1 Linear frequency analysis To analyze the variation of linear frequency of two transverse modes of a fixed-fixed microbeam we numerically solve the nonlinear static equations (given in Appendix

17 Table 1 Dimensions material properties and the electrostatic force coefficients in a fixed-fixed microbeam Quantity Symbol Fixed-fixed beam Length L 500 µm Width B 4 µm Height H 00 nm Side gap g 0 g 1 4.5 µm 7 µm Bottom gap d 500 µm Youngs modulus E.58 10 10 N/m Initial tension N 0 38.336 µn Density ρ 37.4 kg/m 3 Electric constant ɛ 0 8.854 10 1 F/m Fringing coefficients k 1 k k 3 0.945.6 1.3 Fig. (a) Variation of static deflection q 1 verses DC voltage for in-plane mode. (b) Variation of static deflection q verses DC voltage for out-of-plane mode.(c) Variation of in-plane and out-of-plane frequencies with DC voltage for equal interbeam gaps ((g 0 = g 0 = 4.5µm) and unequal interbeam gaps (g 0 = 4.5µm g 1 = 7 µm) and their comparison with experimental results. It also shows 1:1 internal resonance at V dc = 81 V. A) and linear modal dynamic equations given by Eqns. (.19) and (.0) as described in. Subsequently we obtain linear frequencies corresponding to both the modes from Eqn. (.1). Figures (a) and (b) show the variation of static deflection verses DC voltage for in-plane and out-plane modes with a pull-in voltage of about V dc = 199 V. Figure (c) shows the variation of in-plane and out-of-plane linear frequencies verses DC voltage and their comparison with experiments from. As the DC voltage is varied from 0 to 90 V the in-plane frequency decreases due to electrostatic softening effect and the out-of-plane frequency increases due to stretching of the beam in the in-plane direction. Consequently the two frequencies come near to each other and they eventually show 1:1 internal resonance at DC voltage of 81 V. We define this point as coupling point or region. It also shows the variation in-plane and out-of-plane frequencies with DC voltage for equal interbeam gaps (g 0 = g 0 = 4.5µm) with no coupling. In the following section we apply the method of multiple scales to find nonlinear frequency response near the coupling region. We also compare the nonlinear response of different modes when the operating linear frequencies are below the coupling range.

18 Fig. 3 (a) Uncoupled frequency verses Ω ω 1 showing parametric response for in-plane mode. (b) Uncoupled nonlinear frequency verses Ω ω showing duffing response for out-of-plane mode. Here LP denotes limit point. 4. Nonlinear response of uncoupled in-plane and out-of-plane modes At very low DC voltage the linear frequencies of in-plane and out-of-plane modes do not show any coupling region. To find the nonlinear response of uncoupled modes much below the coupling region we neglect the coupling terms from Eqns. (.17) and (.18) and solve the governing equations of each modes separately. To solve the non-linear dynamic equation we consider only the dynamic component as the static deflection is negligible at low DC voltage. Subsequently the method of multiple scales can be used to obtain the modulation equations for each modes separately. While the nonlinear response of in-plane mode turns out to be purely parametric the nonlinear response of out-of-plane mode shows Duffing like response under the influence of direct and parametric forces 15. Figure 3 shows uncoupled nonlinear frequency verses Ω ω 1 showing parametric response for in-plane mode in Fig. 3(a) when V ac = 0.07 V V dc = 0.05 V and Q 1 = 5 10 5. Figure 3(b) shows uncoupled nonlinear response verses Ω ω for the out-of-plane mode when V ac = 5 V V dc = V and Q 1 = 5 10 5. The values of AC and DC voltages are selected to show bi-stability region in the nonlinear response. Now we present coupled nonlinear frequency response of two modes near the coupling region using the methods of multiple scales presented in the theoretical section. 4.3 Validation of MMS solution near coupling region To validate the solutions obtained by solving the modulation Eqns. (3.49) (3.50) (3.51) and (3.5) from the method of multiple scales (MMS) near the coupling region we solve the original modal dynamic Eqns. (.17) and (.18) using the Runge-Kutte method. To compare the results we convert a 1 and a appearing in the modulation equations to equivalent expression of P 1 (t) and P (t) as given by

19 Fig. 4 (a) Comparison of numerical results with solutions based on MMS for in-plane mode at coupling point (b) Comparison of numerical results with solutions based on MMS for outof-plane mode at coupling point. Here we take V dc = 81 V V ac = 0.05 V Q 1 = 5 10 4 Q = 10 6 σ = 0.0858 respectively. Fig. 5 Long-time histories at coupling point for V dc = 81 V V ac = 0.05 V Q 1 = 5 10 4 Q = 10 6 when σ 1 = 0.1 and σ = 0.0858. Eqns. (3.53) and (3.54). Thus P 1 (t) and P (t) obtained from MMS are compared with the solutions obtained from the original equations. Figures 4(a) and (b) show comparisons between numerical results and the solutions based on MMS for inplane and out-of-plane modes near the coupling point for the parameter values V dc = 81 V V ac = 0.05 V Q 1 = 5 10 4 Q = 10 6 and σ = 0.0858. Figures 5(a) and (b) show the long-time histories of the response for in-plane and out-of-plane modes when σ 1 = 0.1 and σ = 0.0.0858. The time histories show that the steady state response of in-plane and out-of-plane modes consists of single frequency.

0 Fig. 6 (a) Comparison of numerical results with solutions based on MMS for in-plane mode at coupling point (b) Comparison of numerical results with solutions based on MMS for outof-plane mode at coupling point. Here we take V dc = 81 V V ac = 0.0009 V Q 1 = 5 10 5 Q = 10 8 σ = 0.0858 respectively. Fig. 7 Long-time histories at coupling point for V dc = 81 V V ac = 0.0009 V Q 1 = 5 10 5 Q = 10 8 when σ 1 = 0.1 and σ = 0.0858. Similarly the comparisons between the numerical results and the solutions based on MMS for in-plane and out-of-plane modes at coupling point (V dc = 81 V and V ac = 0.0009 V) at different quality factors Q 1 = 5 10 5 and Q = 10 8 are shown in Figs. 6(a) and (b). In this case both in-plane and out-of-plane frequency response show two peaks. The long-time histories as shown in Figs. 7(a) and (b) for σ 1 = 0.1 and σ = 0.0858 also show that the steady state response of both in-plane and out-of-plane modes consist of two frequencies corresponding to two peaks appearing in the response. Thus it shows clearly the influence of one mode on another.

1 Fig. 8 (a) Frequency response for different AC voltages below coupling point corresponding to (a) in-plane mode (b) out-of-plane mode. Here V dc = 70 V Q 1 = 5 10 4 Q = 10 6 σ = 1.069. LP denotes limit point. 4.4 Nonlinear response near and below the coupling region Fig. 9 Frequency response for different AC voltages at coupling point corresponding to (a) in-plane mode (b) out-of-plane mode. Here V dc = 81 V Q 1 = 5 10 4 Q = 10 6 σ = 0.0858. LP denotes limit point and H denotes hopf bifurcation point. In this section to show the influence of coupling on nonlinear response near and below the coupling region we analyze the variation of a 1 and a corresponding to in-plane and out-of-plane modes. For the linear frequency relation below the coupling region at V dc = 70 V and near the coupling region at V dc = 81 V we take Q 1 = 5 10 4 Q = 10 6. Figures 8(a) and (b) show the frequency response for different AC voltages below coupling point along in-plane and out-of-plane directions. With the increase in V ac from 0.04 to 0.06 the response amplitude increases

Fig. 10 Frequency response for different AC voltages at coupling point corresponding to (a) in-plane mode (b) out-of-plane mode. Here V dc = 81 V Q 1 = 5 10 5 Q = 10 8 σ = 0.0858. LP denotes limit point. Fig. 11 Frequency response below coupling point corresponding to (a) in-plane mode (b) out-of-plane mode. Here V dc = 79 V V ac = 0.001 V Q 1 = 5 10 4 Q = 10 6 σ = 0.5571538. LP denotes limit point. in both the cases. However only single peak is observed in both the cases due to relatively low quality factor. The frequency response for in-plane motion is found to be linear whereas out-of-plane motion shows nonlinear response. By operating the beam near the coupling region at V dc = 81 V the nonlinear coupled response of two modes shows combined effect of parametric and Duffing like response when the quality factors remains same as Q 1 = 5 10 4 Q = 10 6 which are shown in Figures 9 (a) and (b). The coupling between parametric and duffing response at coupling point is due to simultaneous parametric and direct excitation of microbeam by two symmetrically placed side electrodes and a bottom electrode 15. It is also observed that as AC voltage V ac is increased from 0.05 to 0.14 V the

3 Fig. 1 Frequency response above coupling point corresponding to (a) in-plane mode (b) outof-plane mode. Here V dc = 83 V V ac = 0.06 V Q 1 = 5 10 4 Q = 10 6 σ = 0.1836534. LP denotes limit point. response amplitude and the bandwidth gradually increase with increasing hardening effect. To see the influence of quality factor on the non-linear coupled response near the coupling region we take another set of quality factors Q 1 = 5 10 5 and Q = 10 8 as shown in Figs. 10 (a) and (b). It is observed that coupled response shows two peaks thus clearly indicating the influence of one mode on another. With further increase in AC voltage V ac from 0.0005 to 0.00 V the response amplitude of two modes increase gradually along both directions and frequency response of one of the two modes becomes nonlinear showing hardening effect when V ac is more than 0.0009 V. Figures 11(a) and (b) show the frequency response of the in-plane and out-of-plane directions below the coupling point at DC voltage V dc = 79 V when V ac = 0.001 V Q 1 = 5 10 4 and Q = 10 6. Similarly the frequency response along the in-plane and out-of-plane directions above the coupling point at a DC voltage of V dc = 83 V for V ac = 0.006 V and same values of quality factors are shown in Figures 1(a) and (b). The results in both the cases show that the coupled effect reduces drastically as we go up or below the coupled region. Finally we state that the tuning of nonlinear frequency response of two modes near and below the coupling region by the application DC voltage and quality factors. The study presented in this paper can also be extended to understand the coupling of different modes of beams in MEMS arrays. 5 Conclusion In this paper we have developed a theoretical model for in-plane and out-of-plane motions of a fixed-fixed microbeam separated from two symmetrically placed side electrodes and a bottom electrode. Using the electrostatic force model based on the direct and fringing forces we obtain the partial differential equations governing the nonlinear motion of in-plane and out-of-plane motions. To do linear and nonlinear analysis we obtain the reduced order form of the equations using the Galerkin s

4 method. To analyze the variation of two modes at different DC voltage we plot linear frequencies versus DC voltage. We found that the two modes shows coupling at around DC voltage of 81 V. Thus we obtain 1:1 internal resonance condition near the coupling region. To find the nonlinear response near and below the coupling region we apply the method of multiple scales (MMS). After validating the solution from MMS with numerical solution near the coupling region we analyze the influence of ac voltage and quality factor on the nonlinear response at and near the coupling point. We found that the nonlinear response below the coupling point shows uncoupled response of each modes the response near the coupling region shows different types of coupled response at different quality factors. Acknowledgement This research is supported in part by the Council of Scientific and Industrial Research (CSIR) India ((0696)/15/EMR-II). Appendix A Nonlinear static equations 758.48 q 1 7 α 1 + (10.84 r 1 α 1 10.84 α 1 ) q 1 6 + ( 347.07 α 1 + 347.07 r 1 α 1 + 758.48α q + 61.66 N + 1934.0 1388.9 r 1 α 1 ) q1 5 + ( 654.13 r 1 470.9 r 1 α 1 654.13 10.84α q + 83.15 r 1 N + 10.84 r 1 α q + 470.9 r 1 α 1 83.15 N ) q 1 4 + ( 347.07r 1 α q + 97.00 + 8.1 N 3708.01 r 1 + 97.00 r 1 +151.34 r 1 α 1 11.85 r 1 N 1388.9r 1 α q + 347.07 α q + 8.1 r 1 N ) q 1 3 + ( 38.8 r 1 N 1.33 β s V 10 + 1330.88 r 1 470.9 r 1 α q + 470.9 r 1 α q 1330.88 r 1 38.8 r 1 N + 1.33 β s V 1 ) q 1 + ( 1.30 r 1 N.00 β s V 1.00 β s V 10 r 1 + 500.56 r 1 + 151.34 r 1 α q ) q 1 + 0.83 β s V 1 0.83 β s V 10 r 1 = 0 347.07 q 5 α 3 α 470.9 q 4 α 3 α + ( 151.34 α 3 α 1.85 V g β 3 g + 97.0 + 347.07α 3 α 1 q 1 + 1.85 V 10 α g + 1.85 V 1 α g + 8.1 α 3 N ) q 3 + (.66 V 1 α g 38.8 α 3 N.66 V 10 α g 1330.88 + 1.33 V 10 α g + 1.33 V 1 α g +3.99V g β 3 g 1.33 V g β g 470.9 α 3 α 1 q 1 ) q + (.0 V 1 α g + 500.56 +151.34α 3 α 1 q 1 3.0 V g β 3 g +.0 V g β g + 1.30 α 3 N + 1.0 V 1 α g +1.0 V 10 α g.0 V 10 α g ) q 0.831 V g β g 0.831 V g β g In-plane equation coefficients +0.831 V g β 3 g + 0.831 V 10 α g + 0.831 V 1 α g = 0 m 1i = 5.56 q 1 r 1 3 11.59 q 1 4 r 1 17.11 q 1 5 r 1 + 3.98 r 1 q 1.65 q 1 3 r 1 3 + 3.86 r 1 q 1 3 3.98 q 1 r 1 3 + 34.77 q 1 4 r 1 16.67 q 1 r 1 3.86 q 1 3 r 1 + 5.55 r 1 q 1 + r 1 3 +.65 q 1 3 8.49 q 1 6 11.59 q 1 4 + 17.11 q 1 5

5 k 1i = 511.4 q 1 3 r 1 3 α q + 119.14 r 1 q 1 3 α 1 1534.6q 1 5 r 1 3 α 1 11.11β s V 1 q 1 + 151.34 r 1 3 α q + 76.0 q 1 5 N 686.314 q 1 6 r 1 α 1 + 13808.31 q 1 5 r 1 α 1 41.57 q 1 3 r 1 3 N β s V 1 11.11 β s V 10 r 1 q 1 + 554.90 q 1 4 r 1 N + 41.57 q 1 3 N + 374.15 r 1 q 1 3 N + 500.56 r 1 3 580.07 q 1 4 + 8563.10 q 1 5 53.9 q 1 r 1 N + 137.064733 q 1 3 45.39q 1 6 + 781.01q 1 r 1 3 + 57.4 r 1 q 1 N + 511.4 q 1 3 α q 580.07 q 1 4 r 1 8563.10 q 1 5 r 1 + 1996.33r 1 q 1 137.06 q 1 3 r 1 3 + 11943.58 r 1 q 1 3 1996.33 q 1 r 1 3 + 17406.1 q 1 4 r 1 8343.0 q 1 r 1 11943.58 q 1 3 r 1 + 781.01 r 1 q 1 + 3397.83 q 1 5 α q 313.66 q 1 r 1 α q + 686.31 q 1 4 r 1 α q 184.97 q 1 4 N 138.3 q 1 6 N + 1.31 r 1 3 N 686.31 q 1 6 α 1 + 10193.49 q 1 7 α 1 5104.69 q 1 8 α 1 + 1534.6 q 1 5 α 1 + 454.01q 1 r 1 3 α 1 13808.31q 1 5 r 1 α 1 10193.49q 1 7 r 1 α 1 + 313.66 q 1 4 r 1 3 α 1 + 7.98 β s V 1 q 1 + 84.64 q 1 r 1 3 N + 313.66q 1 4 r 1 α 1 76.1 q 1 5 r 1 N 119.14 q 1 3 r 1 3 α 1 + 5.30 β s V 1 q 1 3 + 0478.94 q 1 6 r 1 α 1 5.30 β s V 10 q 1 3 374.15 q 1 3 r 1 N 57.4 q 1 r 1 3 N β s V 10 r 1 3 9370.99 q 1 4 r 1 α 1 75.44 q 1 4 α q 1701.56 q 1 6 α q + 84.64 r 1 q 1 N + 706.38 r 1 q 1 α q 7.97 β s V 10 r 1 q 1 706.38 q 1 r 1 3 α q 460.77 q 1 3 r 1 α q + 1041. q 1 r 1 3 α q + 460.77 r 1 q 1 3 α q 3397.83 q 1 5 r 1 α q 184.97 q 1 4 r 1 N 75.44 q 1 4 r 1 α q + 1041. r 1 q 1 α q n 1i = 1041. q 1 r 1 3 α 1 + 511.4 q 1 3 α 1 706.38 q 1 r 1 3 α 1 + 151.33 r 1 3 α 1 313.66 r 1 q 1 α 1 + 706.38 r 1 q 1 α 1 + 686.31 q 1 4 r 1 α 1 75.44 q 1 4 α 1 511.4 q 1 3 r 1 3 α 1 75.44 q 1 4 r 1 α 1 1701.56 q 1 6 α 1 + 1041. r 1 q 1 α 1 + 7460.77r 1 q 1 3 α 1 3397.83q 1 5 r 1 α 1 + 3397.83q 1 5 α 1 460.77q 1 3 r 1 α 1 n i = 686.31q 1 5 r 1 α 1 5104.69q 1 7 α 1 13808.31q 1 4 r 1 α 1 9370.99r 1 q 1 3 α 1 + 1534.6 q 1 4 α 1 + 10193.49 q 1 6 α 1 + 13808.31 q 1 4 r 1 α 1 + 454.01 q 1 r 1 3 α 1 1534.6 q 1 4 r 1 3 α 1 + 313.66 q 1 3 r 1 3 α 1 686.31 q 1 5 α 1 119.14 q 1 r 1 3 α 1 + 313.66 q 1 3 r 1 α 1 + 0478.94 q 1 5 r 1 α 1 10193.49 q 1 6 r 1 α 1 + 119.14 r 1 q 1 α 1

6 n 3i = 460.77 q 1 3 r 1 α 75.44 q 1 4 r 1 α 313.66 q 1 r 1 α + 460.77 r 1 q 1 3 α 3397.83q 1 5 r 1 α 1701.56q 1 6 α + 706.38r 1 q 1 α + 511.4q 1 3 α + 151.34 r 1 3 α + 686.31q 1 4 r 1 α + 1041.q 1 r 1 3 α + 1041.r 1 q 1 α 511.4q 1 3 r 1 3 α 706.38q 1 r 1 3 α 75.44q 1 4 α + 3397.83q 1 5 α n 4i = 4550.87q 1 4 α q 10.84q 1 3 r 1 3 α q 4550.87q 1 4 r 1 α q + 1365.63 q 1 4 r 1 α q + 905.54 q 1 3 r 1 α q 6795.66 q 1 5 r 1 α q 905.54 q 1 3 r 1 α q + 141.76 r 1 q 1 α q 141.76 q 1 r 1 3 α q 647.33 r 1 q 1 α q + 08.44 r 1 q 1 α q + 08.44 q 1 r 1 3 α q + 10.84 q 1 3 α q + 30.67 r 1 3 α q + 6795.66 q 1 5 α q 3403.13 q 1 6 α q n 5i =.0 β s r 1 3 7.98 β s r 1 q 1 11.11 β s r 1 q 1 5.30 β s q 1 3 n 6i = 3.70 β s q 1 3 1.66 β s + 6.0 β s q 1 7.98 β s q 1 n 7i = 1041.q 1 3 r 1 α 706.38q 1 r 1 3 α 75.44q 1 5 r 1 α + 151.33 q 1 r 1 3 α 3397.83 q 1 6 r 1 α 460.77 q 1 4 r 1 α + 3397.83 q 1 6 α + 511.4 q 1 4 α + 1041. q 1 3 r 1 3 α 1701.56 q 1 7 α 313.66 r 1 q 1 3 α + 460.77 q 1 4 r 1 α 511.4q 1 4 r 1 3 α 75.44q 1 5 α + 686.31q 1 5 r 1 α + 706.38q 1 r 1 α n 8i = 6795.66q 1 6 r 1 α q 905.54q 1 4 r 1 α q 141.76q 1 r 1 3 α q 647.33q 1 3 r 1 α q 4550.87q 1 5 r 1 α q + 08.44q 1 3 r 1 3 α q + 141.76r 1 q 1 α q + 30.67 q 1 r 1 3 α q + 10.84 q 1 4 α q 10.84 q 1 4 r 1 3 α q + 6795.66 q 1 6 α q 4550.87 q 1 5 α q + 08.44 q 1 3 r 1 α q + 1365.63 q 1 5 r 1 α q 3403.13 q 1 7 α q + 905.54 q 1 4 r 1 α q n 9i = 3.98 β s r 1 q 1 +.65 β s q 1 4 + 3.98 β s r 1 q 1 + 5.55 β s r 1 q 1 3 + β s r 1 3 q 1 1.85 β s q 1 3 0.83 β s r 1 3 3.0 β s r 1 q 1 n 10i = 3.0 β s q 1.49 β s r 1 q 1 1.33 β s r 1 q 1 3 + β s r 1 1.85 β s q 1 4 + 3.98 β s q 1 3 + 0.83 β s q 1 + 3.0 β s r 1 q 1 n 11i = 8.49 q 1 6 c 1 11.59 q 1 4 r 1 c 1 + 3.98 r 1 q 1 c 1 + 5.55 q 1 r 1 3 c 1 + 5.55 r 1 q 1 c 1 3.98 q 1 r 1 3 c 1 3.86 q 1 3 r 1 c 1 + 17.11 q 1 5 c 1 + 3.86 q 1 3 r 1 c 1 +.65 q 1 3 c 1 16.67 r 1 q 1 c 1 11.59 q 1 4 c 1 17.11 q 1 5 r 1 c 1.65 q 1 3 r 1 3 c 1 + r 1 3 c 1 + 34.77 q 1 4 r 1 c 1 k 1i λ 1 = m 1i t 6 = n 6i m 1i t 1 = n 1i m 1i t 7 = n 7i m 1i t = n i m 1i t 8 = n 8i m 1i t 3 = n 3i m 1i t 9 = n 9i m 1i t 4 = n 4i m 1i t 10 = n 10i m 1i t 5 = n 5i m 1i t 11 = n 11i m 1i