Question

Λ steel ball weighing 64lb stretches a spring by 4ft when suspended vertically to reach its equilibrium position. If, at time t=0, the ball is displaced 6 in below its equilibrium position and released, what will the position y=y(t) (positive downwards) of the ball be at time t (sec)? If the ball will execute pure oscillations, find the amplitude, frequency and period of the oscillations. How fast will the ball passes its equilibrium position each time? Assume that friction may be neglected.

Answer 1

A steel ball weighing 64lb stretches a spring by 4ft. The amplitude, frequency, and period of the oscillations can be calculated using the provided information. The ball passes its equilibrium position each time it completes an oscillation.

Step-by-step explanation:

Given that a steel ball weighing 64lb stretches a spring by 4ft when suspended vertically to reach its equilibrium position, we can find the spring constant(k) using Hooke's Law:

F = -kx

Where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.

So, k = F / x = (64lb * 32.2ft/s^2) / 4ft = 515.2 lb-ft/s^2.

Since the ball is displaced 6in below its equilibrium position, we can calculate the amplitude(A) of the oscillations:

A = 6in = 0.5ft.

The frequency(f) of the oscillations can be calculated using the formula:

f = 1 / (2π) * (√(k/m))

Where m is the mass of the ball. Assuming the mass of the ball is 64 lb / (32.2 ft/s^2) = 2 lb-s^2/ft, the frequency equals 3.89 Hz.

The period(T) of the oscillations is the reciprocal of the frequency, so T = 1 / f ≈ 0.26s.

The ball passes its equilibrium position each time it completes an oscillation, so the distance it passes is equal to 2A, which is 1ft.