On a horizontal frictionless table an open topped 520...

On a horizontal, frictionless table, an open-topped $5.20-\mathrm{kg}$ box is attached to an ideal horizontal spring having force constant 375 $\mathrm{N} / \mathrm{m}$ . Inside the box is a 3.44 -kg stone. The system is oscillating with an amplitude of 7.50 $\mathrm{cm} .$ When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box. Find (a) the period and (b) the amplitude of the resulting motion of the box. (c) Without doing any calculations, is the new period greater or smaller than the original period? How do you know?

Transcript

00:01 Some reminders about simple harmonic motion.

00:04 So first of all, there is a force on an object due to a spring that's proportional to the object's displacement from equilibrium.

00:17 That force we can write with newton's second law is equal to the mass of the object times its acceleration.

00:26 So a result of this type of force is that the object oscillates with an angular frequency given by the square root of k, the spring constant, s, over the mass of the object.

00:45 Furthermore, if there is no friction, there is energy conservation, such that your e -total is a sum of kinetic.

00:58 Energy plus potential energy in the elastic spring.

01:04 We can furthermore break out the kinetic energy as one -half mv squared, and the elastic potential energy is one -half k -x squared, x being the position of the mass relative to its equilibrium position.

01:24 So we can think also about a mass on a spring as having a continually oscillating energy that remains constant but trades off the kinetic and the potential energies.

01:40 As an example of such a situation, let us pretend that there is a box oscillating on a spring without friction.

01:51 We know the spring constant.

01:53 We know the mass of the box, but there is some object in the box with an appreciable mass that you can pluck out whenever you're you want to.

02:03 So this is not a closed system, meaning that the energy can change.

02:14 And in such a system, we are going to pretend that we pluck the mass out when it is going through the little red object inside the box when the box is at equilibrium.

02:38 And the question is, what happens to the the subsequent motion.

02:48 Well, first of all, we can see that as the mass decreases and the spring force remains pretty much the same, so we can use kx equals m times a.

03:04 If the mass goes down and the other side of the equation remains constant, that means the acceleration must go up.

03:17 And that says that the box must turn around much quicker, and we would expect the period to be quicker or to go down as a result of that.

03:33 So we could calculate the new period.

03:40 We'll call it period prime after the object gets plucked out of the box.

03:47 The angular frequency is 2 pi of the period, that new period, and we are going to calculate it with the spring constant square root, as well as the mass of the box that's left over.

04:06 We're calling that a big m.

04:08 We'll show the little m as a red mass when we need it.

04:11 So we can certainly calculate the new period is equal to 2 pi square root.

04:20 Of mass over the spring constant.

04:26 And let's figure that out to pi.

04:31 The box by itself is 5 .20 kilograms.

04:37 The spring constant is 375 newtons per meter...