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A vibration platform oscillates up and down with a fixed amplitude of 9.7 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?

Sagot :

Answer:

1.6 Hz

Explanation:

In simple harmonic motion, acceleration is given by the formula;

a = Aω²

Where;

A is amplitude

ω is angular velocity.

We are given A = 9.7 cm = 0.097 m

We know that acceleration due to gravity is 9.8 m/s².

Thus;

9.8 = 0.097ω²

ω² = 9.8/0.097

ω = √101.031

ω = 10.05 rad/s

Formula for frequency is;

f = ω/2π

Thus;

f = 10.05/2π

f = 1.6 Hz

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