Simple harmonic motion involving circuit

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SUMMARY

The discussion centers on calculating the maximum frequency at which a small body remains in contact with a diaphragm of a loudspeaker undergoing simple harmonic motion (SHM). The diaphragm oscillates with an amplitude of 10 µm, and the maximum frequency is determined to be 498.24 Hz. This frequency is derived by equating the maximum acceleration of the diaphragm to the acceleration due to gravity (9.8 m/s²). The calculations utilize the equations of motion for SHM, specifically x = A cos(wt - φ), and address concerns regarding the treatment of phase angle and the direction of acceleration.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM) principles
  • Familiarity with the equations of motion: x = A cos(wt - φ), v = -wA sin(wt - φ), a = -w²A cos(wt - φ)
  • Basic knowledge of oscillatory systems and forces
  • Concept of phase angle in oscillatory motion
NEXT STEPS
  • Study the derivation of maximum acceleration in simple harmonic motion
  • Learn about the effects of phase angle on oscillatory motion
  • Explore applications of SHM in real-world systems, such as loudspeakers
  • Investigate the relationship between frequency, amplitude, and energy in oscillatory systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillations, as well as educators seeking to enhance their understanding of simple harmonic motion in practical applications.

Eric_meyers
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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??
 
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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.
 

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