Question

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.

Answer 1

Answer:

 f = 2ππ √g / A

Explanation:

The expression that describes the simple harmonic movement is

              x = A cos (wt + Ф)

So that the penny should the surface when the system is going down the system must in faster than the acceleration of gravity, let's look for this acceleration

              v = dx / dt

             a = dv / dt = d²x / dt²

              .a = - A w² cos (w t + Ф)

The maximum value of this acceleration is when the cosine value ±1

            a =  A w²

For the penny to leave the surface the limit acceleration is

          a = g

           

          g = A w²

          w² = g/A

          w = 2π f

          g / A = 4 π² f²

           f = 2ππ √g /A

For values ​​greater than this frequency the penny separates from the surface in the downward movement