Vibration Platform: Find Frequency for Rock Clatter

[gonzalo12345]

Homework Statement

A vibration platform oscillates up and down with an amplitude of 10.4 cm at a controlled variable frequency. Suppose 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?

Homework Equations

I need to know how to start this problem and what equations to use

The Attempt at a Solution

 

[tiny-tim]

A vibration platform oscillates up and down with an amplitude of 10.4 cm at a controlled variable frequency. Suppose 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?

Hi gonzalo12345!

This is like a roller-coaster problem … the rock will lose contact when the platform is accelerating downward so fast that the rock cannot fall fast enough to keep up!

So work out the maximum acceleration of the platform (it will depend on frequency, obviously).

 

[baba89]

To start this problem, we need to understand the concept of resonance and how it relates to the vibration platform and the rock. Resonance occurs when an object is subjected to a periodic force at a frequency equal to its natural frequency. In this case, the vibration platform is providing the periodic force and the rock has a natural frequency at which it will begin to clatter.

To find the frequency at which the rock will begin to clatter, we can use the equation for natural frequency:

f = 1/2π √(k/m)

Where f is the frequency, k is the spring constant of the vibration platform, and m is the mass of the rock.

Since the mass of the rock is unknown, we can use the amplitude of the vibration platform to determine the spring constant. The equation for spring constant is:

k = mg/A

Where g is the acceleration due to gravity (9.8 m/s^2) and A is the amplitude (10.4 cm = 0.104 m).

Substituting the values into the first equation, we get:

f = 1/2π √(mg/Am)

Simplifying, we get:

f = 1/2π √(g/A)

Plugging in the values for g and A, we get:

f = 1/2π √(9.8/0.104)

Solving for f, we get a frequency of approximately 5.32 Hz.

Therefore, the rock will just begin to leave the surface and clatter at a frequency of 5.32 Hz.