Simple Harmonic Motion and frequency

[andyatk14]

Homework Statement

A mass sits on a platform undergoing simple harmonic motion in the vertical
direction. The mass loses contact with the platform when the amplitude exceeds
6.2 cm. What is the frequency, f (in Hz), of the platform’s vibration? Take g =
9.8ms−2.

Homework Equations

The Attempt at a Solution

I'm not sure where to start with this.

 

[gneill]

What do you know about SHM, and in particular, the relationships between position, velocity, and acceleration? How about bodies in free-fall? What do you know about them?

 

[andyatk14]

Thank you for your reply. I have an answer for the frequency of 2Hz which I got by saying the maximum value of the acceleration was equal to g, which is equal to omega^2*A, where A is the amplitude. I believe that may be right.

 

[gneill]

Thank you for your reply. I have an answer for the frequency of 2Hz which I got by saying the maximum value of the acceleration was equal to g, which is equal to omega^2*A, where A is the amplitude. I believe that may be right.

If you can justify why maximum acceleration should equal g then I'd say you're done.

 

[rwisz]

I would approach this problem by first understanding the concept of simple harmonic motion. Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from the equilibrium position, and the motion is sinusoidal in nature. This type of motion is commonly observed in systems such as springs, pendulums, and mass-spring systems.

In this problem, we are given a mass undergoing simple harmonic motion on a platform. We are also given the maximum amplitude of the motion, which is when the mass loses contact with the platform. This means that the platform's vibration is strong enough to overcome the gravitational force acting on the mass.

To calculate the frequency of the platform's vibration, we can use the equation for the period of simple harmonic motion, T = 1/f, where T is the time it takes for one complete cycle of the motion and f is the frequency. We can also use the equation for the maximum amplitude of simple harmonic motion, A = ω^2/g, where A is the maximum amplitude, ω is the angular frequency, and g is the acceleration due to gravity.

Substituting the given values, A = 6.2 cm = 0.062 m and g = 9.8 m/s^2 into the equation A = ω^2/g, we can solve for ω. This gives us ω = √(Ag) = √(0.062*9.8) = 0.248 rad/s.

Next, we can use the equation T = 1/f to solve for the period, T. Since the period is the time it takes for one complete cycle, we can assume that when the mass loses contact with the platform, it has completed one full cycle. Therefore, the period is equal to the time it takes for the mass to reach its maximum amplitude. This can be calculated using the equation T = 2π/ω, where π is the constant pi.

Substituting the value of ω, we get T = 2π/0.248 = 25.48 s.

Finally, we can use the equation f = 1/T to calculate the frequency, f. Substituting the value of T, we get f = 1/25.48 = 0.039 Hz.

Therefore, the frequency of the platform's vibration is 0.039 Hz.