This is an Oscillations question.
A horizontal plate is vibrating vertically in simple harmonic motion at a frequency of 20Hz. What is the maximum amplitude of vibration so that a coin of 0.0080kg resting on the plate always remains in contact with it?
Solve
To find the maximum amplitude of vibration, we need to consider the gravitational force acting on the coin and equate it to the maximum centripetal force.
1. The gravitational force acting on the coin is given by the formula: F_gravity = m * g, where m is the mass of the coin and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The centripetal force required to keep the coin in contact with the plate is given by the formula: F_centripetal = m * ω^2 * A, where m is the mass of the coin, ω is the angular frequency (2πf), and A is the amplitude of vibration.
3. Set the gravitational force equal to the centripetal force and solve for A:
m * g = m * ω^2 * A
4. Since the mass of the coin cancels out, we can simplify the equation to:
g = ω^2 * A
5. Substituting the given values, the angular frequency ω can be calculated using the formula: ω = 2πf.
ω = 2π * 20 Hz
6. Calculate ω^2 and substitute it back into the equation:
A = g / ω^2
A = (9.8 m/s^2) / ( (2π * 20 Hz)^2 )
7. Calculate the final value for A:
A ≈ 0.0314 m (rounded to four significant figures)
Therefore, the maximum amplitude of vibration should be approximately 0.0314 meters (or 31.4 mm) to ensure that the coin remains in contact with the plate.