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Supplementary material
Supplementary Figure 1
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Frequency - kHz
Supplementary Figure 1: Average coherence between microphone signal
and mechanical response across the locust tympanum.
Supplementary Figure 2
Supplementary Figure 2: Extensive deposition of material due to the evaporation of liquids during FIB milling.
Supplementary Figure 3
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Supplementary Figure 3: Frequency response of locust membrane to acoustic stimulus.
Supplementary Note 1 - Derivation of wave equations
Membrane wave equation
For a membrane in tension, T, in the x-y plane, the wave equation for small-amplitude transverse
waves are derived by considering a square element Δx Δy of the membrane. When the membrane is
displaced, there arises a force in the z direction from each of the two pairs of tensile forces T Δx and T
Δy acting on the four edges of the square element. If we consider strips of the membrane of width Δx
extending in the y direction then the net force for this strip is:
 z 
 z  
2z
Tx  
     Tx 2 y
y
 y  y  y  y  y 
If a strip of width Δy extending in the x direction is also consider we get:
 z 
2z
 z  
Ty  
     T y 2 x
x
 x  x  x  x  x 
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Their sum equals the mass ρΔxΔy of the element times its acceleration:
2 z 2 z 
2z
xyT  2  2   xy 2
y 
t
 x
By dividing out the Δx Δy we gain the wave equation:
2 z 2 z  2 z


x 2 y 2 T t 2
By inserting a solution,
z (t )  Z 0 e i ( k ( x  y ) t )
into the wave equation we get:
kx  ky 
2
2

T
2
Arranging to:
  kx2  ky2
T

Since:
~
2
2
kx  k y  k
Velocity of transverse flexural wave is given by:
v

k

T


T
h
Stiff plate wave equation
A stiff plate is a plane layer without tension but with non-negligible stiffness. For our purpose the
plate will assumed to be thin (thickness is small in comparison to its length and breadth). The wave
equation for transverse “bending waves” of small displacement is derived (by equating the transverse
force accompanying the plate deformation, given by the Sophie-Germain equation.
D(2 z) 2
With the inertia of the plate per unit area:
h 2 z / t 2
2 z G

( 2 ) 2  0
t
h
Where G:
Eh 3
G
12 1   2


is the flexural stiffness of the plate. E is the Young’s modulus and ν is the Poisson ratio. In the case of a
one-dimensional wave in the x direction we obtain:
2z G 4z

0
t
h x 4
which is a fourth order in x, and second order in t. A suitable solution must have sinusoidal time
dependence:
z ( x, t )  f ( x )e  iwt
Substituting this trial solution into (10), we find f(x) must satisfy:
2 f 
G 4 f
0
h x 4
If the spatial dependence is of the form f(x) = eαx we find:
 
4
h 2
G
 k4
where:
 h 2 

k  
G


1/ 4
which has the four roots α = k, -k, ik, -ik, giving the general solution:
z( x, t )  ( Aekx  Bekx  Ceikx  Deikx )e t
where A, B, C, D are arbitrary constants. Values can be chosen to construct a sinusoidal travelling
flexural wave of form:
z( x, t )  Z 0 cos(kx  t )
Finally the velocity of this travelling flexural wave is then:

1/ 4
 G 2 

v   
k  h 
1/ 4


E

 
2 
 12 (1  ) 
h