# Examples of AFM cantilevers
## For a BS-75kHz

L=225 w=28 t=3 h=17+3/2 (is actually trapezoidal)}
$k_{N}={\frac {1}{4}}Yw\left({\frac {t}{L}}\right)^{3}={\frac {1}{4}}\left(160000\right)28\left({\frac {3}{225}}\right)^{3}=2.\,654\,8$
$f[Hz]={\frac {t\beta _{i}^{2}}{4\pi L^{2}}}{\sqrt {\frac {Y}{3\rho }}}={\frac {\left(3\ast 10^{-6}\right)\left(1.875\right)^{2}}{4\pi \left(225\ast 10^{-6}\right)^{2}}}{\sqrt {\frac {\left(160\ast 10^{9}\right)}{3\ast 2330}}}=79318.Hz$
$2\left({\frac {wh}{tL}}\right)^{2}=2\left({\frac {28\ast 18.5}{3\ast 225}}\right)^{2}=1.\,177\,8$ so $k_{lat}~k_{tor}.$
$k_{tor}=k_{N}{\frac {1}{2}}\left({\frac {L}{h}}\right)^{2}=2.65{\frac {1}{2}}\left({\frac {225}{18.5}}\right)^{2}=195.\,99$
$k_{lat}={\frac {1}{4}}Y{\frac {tw^{3}}{L^{3}}}={\frac {1}{4}}\left(160000\right)3\left({\frac {28}{225}}\right)^{3}=231.\,26$

## For a MPP311

Specs: 13 kHz, 0.45 N/m : L=440 w=30 t=4 h=17.5+2 (is it actually trapezoidal??)
$k_{N}={\frac {1}{4}}Yw\left({\frac {t}{L}}\right)^{3}={\frac {1}{4}}\left(160000\right)30\left({\frac {4}{440}}\right)^{3}=0.901\,58$
$f$ $[Hz]={\frac {t\beta _{i}^{2}}{4\pi L^{2}}}{\sqrt {\frac {Y}{3\rho }}}={\frac {\left(4\ast 10^{-6}\right)\left(1.875\right)^{2}}{4\pi \left(440\ast 10^{-6}\right)^{2}}}{\sqrt {\frac {\left(160\ast 10^{9}\right)}{3\ast 2330}}}=27655.Hz$
$2\left({\frac {wh}{tL}}\right)^{2}=2\left({\frac {30\ast 19.5}{4\ast 440}}\right)^{2}=0.220\,96$ so $k_{lat}<k_{tor}.$
$k_{tor}=k_{N}{\frac {1}{2}}\left({\frac {L}{h}}\right)^{2}=0.9{\frac {1}{2}}\left({\frac {440}{19.5}}\right)^{2}=229.\,11$
$k_{lat}={\frac {1}{4}}Y{\frac {tw^{3}}{L^{3}}}={\frac {1}{4}}\left(160000\right)4\left({\frac {30}{440}}\right)^{3}=50.\,714$

## For a MPP211

Specs: 50 kHz, 1.5 N/m : L=215 w=30 t=4 h=17.5+2 (is it actually trapezoidal??)
$k_{N}={\frac {1}{4}}Yw\left({\frac {t}{L}}\right)^{3}={\frac {1}{4}}\left(160000\right)30\left({\frac {4}{215}}\right)^{3}=7.\,727\,6$
$f$ $[Hz]={\frac {t\beta _{i}^{2}}{4\pi L^{2}}}{\sqrt {\frac {Y}{3\rho }}}={\frac {\left(4\ast 10^{-6}\right)\left(1.875\right)^{2}}{4\pi \left(215\ast 10^{-6}\right)^{2}}}{\sqrt {\frac {\left(160\ast 10^{9}\right)}{3\ast 2330}}}=1.\,158\,2\times 10^{5}Hz$
$2\left({\frac {wh}{tL}}\right)^{2}=2\left({\frac {30\ast 19.5}{4\ast 215}}\right)^{2}=0.925\,43$
so $k_{lat}~k_{tor}.$
$k_{tor}=k_{N}{\frac {1}{2}}\left({\frac {L}{h}}\right)^{2}=7.7{\frac {1}{2}}\left({\frac {215}{19.5}}\right)^{2}=468.\,02$
$k_{lat}={\frac {1}{4}}Y{\frac {tw^{3}}{L^{3}}}={\frac {1}{4}}\left(160000\right)4\left({\frac {30}{215}}\right)^{3}=434.\,68$

## For a MPP111

Specs: 200 kHz, 20 N/m : L=115 w=30 t=4 h=17.5+2 (is it actually trapezoidal??)
$k_{N}={\frac {1}{4}}Yw\left({\frac {t}{L}}\right)^{3}={\frac {1}{4}}\left(160000\right)30\left({\frac {4}{115}}\right)^{3}=50.\,497$
$f$ $[Hz]={\frac {t\beta _{i}^{2}}{4\pi L^{2}}}{\sqrt {\frac {Y}{3\rho }}}={\frac {\left(4\ast 10^{-6}\right)\left(1.875\right)^{2}}{4\pi \left(115\ast 10^{-6}\right)^{2}}}{\sqrt {\frac {\left(160\ast 10^{9}\right)}{3\ast 2330}}}=4.\,048\,4\times 10^{5}Hz$
$2\left({\frac {wh}{tL}}\right)^{2}=2\left({\frac {30\ast 19.5}{4\ast 115}}\right)^{2}=3.\,234\,6$ so $k_{lat}>k_{tor}.$
$k_{tor}=k_{N}{\frac {1}{2}}\left({\frac {L}{h}}\right)^{2}=50.4{\frac {1}{2}}\left({\frac {115}{19.5}}\right)^{2}=876.\,45$
$k_{lat}={\frac {1}{4}}Y{\frac {tw^{3}}{L^{3}}}={\frac {1}{4}}\left(160000\right)4\left({\frac {30}{115}}\right)^{3}=2840.\,5$