# Simple harmonic motion involving circuit

## Homework Statement

"A small body rests on a horizontal diaphragm of a loudspeaker which is
supplied with an alternating current of constant amplitude but variable frequency.
If the diaphragm executes simple harmonic oscillation in the vertical
direction of amplitude 10 µm, at all frequencies, find the greatest frequency
for which the small body stays in contact with the diaphragm."

## Homework Equations

x = A cos (wt - phi)

## The Attempt at a Solution

Ok, so my reasoning here is the mass will stay on the platform so long as the acceleration of the platform does not exceed the acceleration of gravity (the only force holding the mass on the platform)

so, the maximum frequency must be the frequency that makes the acceleration exactly equal to gravity.

x = A cos (wt - phi)

v = -w A sin (wt - phi)

a = - w^2 A cos (wt - phi)

9.8 = -w^2 * 10^-6

w = 3130.49

f = w/(2 * pi) = 498.24 Hz.

ok.. so a lot of things make me uncomfortable in this problem.

First, I'm setting phi = 0 and t to be some value that makes cos = 1 ..I'm not sure why I can do that so that makes me uncomfortable.

Secondly I'm ignoring the minus sign in my equation. That too makes me uncomfortable.

Is there an error in my logic??

## Answers and Replies

rl.bhat
Homework Helper
The acceleration of the diaphragm is not constant. It is maximum at the extreme position of the diaphragm. So cos should be 1.
When the diaphragm is retreating, if its acceleration is less than or equal to g, the small body stays in contact with the diaphragm. The retreating action takes in to account the negative sign.