Answer
$30000$
Work Step by Step
Let's apply equation 10.6 $f=\frac{\omega}{2\pi}$ to find the frequency of the diaphragm.
$f=\frac{\omega}{2\pi}$ ; Let's plug known values into this equation.
$f=\frac{7.54\times10^{4}rad/s}{2\pi rad}=12000\space s^{-1}$
Number of times = $ft=12000\space s^{-1}\times 2.5\space s\approx30000$
So, the number of times the diaphragm moves back and forth in 2.5 s = 30000
