A vibration platform oscillates up and down with an amplitude of 10.2 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?

To determine the frequency at which the rock will just begin to leave the surface and start to clatter, we need to consider the gravitational force acting on the rock and the centripetal force acting on it due to the motion of the vibrating platform.

First, let's determine the acceleration experienced by the rock on the platform. Since the platform is oscillating up and down, the acceleration will vary with time. At the extreme points of the motion (where the platform is at its maximum displacement), the acceleration will be maximum.

The maximum acceleration of the platform can be determined using the amplitude and the frequency of the vibration. In this case, the amplitude is given as 10.2 cm.

The formula to calculate the maximum acceleration (a_max) from the amplitude (A) and frequency (f) is:

a_max = 4π²Af²

Plugging in the values:

a_max = 4 * 3.1416^2 * 0.102m * f²

Now, let's consider the forces acting on the rock. The gravitational force pulling the rock down is given by:

F_gravity = m * g

where m is the mass of the rock and g is the acceleration due to gravity (approximately 9.8 m/s²).

At the point of clattering, the rock just loses contact with the platform. At this instant, the centripetal force acting on the rock must be equal to the gravitational force. The centripetal force is given by:

F_centripetal = m * a_max

Setting these two forces equal to each other:

m * a_max = m * g

Simplifying and rearranging the equation:

a_max = g

Now we have an equation relating the maximum acceleration of the platform to the acceleration due to gravity. This means that the platform's maximum acceleration is equal to the acceleration due to gravity at the point of clattering.

Plugging in the value for a_max from earlier:

4 * 3.1416^2 * 0.102m * f² = 9.8 m/s²

Now, we can solve for the frequency (f):

f² = (9.8 m/s²) / (4 * 3.1416^2 * 0.102m)

f = √((9.8 m/s²) / (4 * 3.1416^2 * 0.102m))

f ≈ 2.52 Hz

Therefore, the frequency at which the rock will just begin to leave the surface and start to clatter is approximately 2.52 Hz.

When max acceleration is greater than g.