A loudspeaker produces a musical sound by the oscillation of a diaphragm. If the amplitude of oscillation is limited to 1.20 × 10^-3 mm, what frequencies will result in the acceleration of the diaphragm exceeding g?

Here the value of maximum acceleration a_x of the diaphragm is equal to the acceleration due to gravity g and which has value 9.81 m/s² on the surface of Earth.
To obtain the frequency of diaphragm, substitute 9.81 m/s² for acceleration ax, 1.20×10^-3 mm for the amplitude of oscillation in the equation f = (√ ax/ xm) /2π,
f = (√ a_x/ x_m) /2π
= (√(9.81 m/s² /1.20×10^-3 mm) (1 mm/10^-3 m )) /2×3.14
=455/s
= (455/s) (1 Hz/(1/sec))
= 455 Hz
From the above observation we conclude that the frequency of diaphragm would be 455 Hz.