Simple harmonic motion involving circuit

[Eric_meyers]

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??

 

[rl.bhat]

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 into account the negative sign.

 

[tomcue]

Your reasoning is correct, but there are a few things to consider.

Firstly, the minus sign in the equation for acceleration is important because it indicates the direction of the acceleration. In this case, the acceleration is directed upwards (opposite to the direction of gravity), so the minus sign is necessary to account for that.

Secondly, setting phi = 0 and cos = 1 is valid because you are assuming that the oscillation is in phase with the alternating current, which means that the diaphragm is at its maximum displacement when the acceleration is equal to gravity. This is the worst case scenario, so it gives you the maximum frequency at which the mass will stay in contact with the diaphragm.

Lastly, it is important to note that this problem assumes ideal conditions, where there is no friction or other external forces acting on the mass. In reality, there will always be some friction present, so the actual maximum frequency may be slightly lower.