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A vibration platform oscillates up and down with a fixed amplitude of 8.1 cm and a controlled frequency that can be varied. If a small rock of unknown mass is placed on the platform, at what frequency will the rock just begin to leave the surface so that it starts to clatter? Hint: You should recall the condition that helps determine when two objects lose contact.

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Answer:

F = 0.02917hz

Explanation:

From the problem we could infer that it involves simple harmonic motion.

For simple harmonic motion, maximum accelration(g) = Amplitude(A)×Angular velocity(w)²

g = Aw²

Simple harnonic motion= A sin(wt)

Therefore, w = √g/A

g = 9.8m/s² and A = 8.1cm

w = √9.8/8.1

w = 1.0999rad/s

Therefore, frequency (f) = w/2Ï€

F = 1.0999/2Ï€

F = 0.02917hz

The frequency will the rock just begin to leave the surface is 0.02917hz

Simple harmonic equation:

Since A vibration platform oscillates up and down with a fixed amplitude of 8.1 cm

Now we know that

The maximum accelration(g) should be

= Amplitude(A)×Angular velocity(w)²

g = Aw²

Here

Simple harmonic motion= A sin(wt)

Now

w = √g/A

Here g = 9.8m/s² and A = 8.1cm

w = √9.8/8.1

w = 1.0999rad/s

Now, frequency (f) = w/2Ï€

F = 1.0999/2Ï€

F = 0.02917hz

learn about rock here: https://brainly.com/question/4822220

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