College Physics (4th Edition)

(a) The maximum force acting on the diaphragm is $1420~N$ (b) The mechanical energy of the diaphragm is $0.128~J$
(a) We can find the maximum acceleration: $a_m = A~\omega^2$ $a_m = A~(2\pi~f)^2$ $a_m = (1.8\times 10^{-4}~m)~(2\pi)^2~(2000~Hz)^2$ $a_m = 28,424~m/s^2$ We can find the maximum force $F_m$: $F_m = m~a_m$ $F_m = (0.050~kg)(28,424~m/s^2)$ $F_m = 1420~N$ The maximum force acting on the diaphragm is $1420~N$ (b) We can find the mechanical energy of the diaphragm: $E = \frac{1}{2}m~v_m^2$ $E = \frac{1}{2}m~(A~\omega)^2$ $E = \frac{1}{2}m~A^2~(2\pi~f)^2$ $E = \frac{1}{2}(0.050~kg)~(1.8\times 10^{-4}~m)^2~(2\pi)^2~(2000~Hz)^2$ $E = 0.128~J$ The mechanical energy of the diaphragm is $0.128~J$.