You have connected a loudspeaker that is facing upwards to a variable frequency source and notice that a small screw has fallen into the loudspeaker. As you increase the frequency of the signal you notice that at a frequency of 20.0 Hz the screw no longer moves with the speaker but starts to leave the surface of the speaker intermittently. What must be the amplitude of the oscillations?
A) 0.621 mm
B) 2.45 mm
C) 1.60 mm
D) 0.893 mm
E) 0.200 mm

Question

Quizplus · Accepted Answer

The screw leaves the surface when the acceleration of the speaker equals gravitational acceleration. Using the formula a = (2πf)^2 * A, where a = 9.8 m/s² and f = 20 Hz, solving for A gives approximately 0.621 mm.

You Have Connected a Loudspeaker That Is Facing Upwards to a Variable