While driving behind a car traveling at 3.00 m / s, you...

While driving behind a car traveling at $3.00 \mathrm{~m} / \mathrm{s},$ you notice that one of the car's tires has a small hemispherical bump on its rim as shown in Figure $\mathrm{P} 15.16$. (a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radii of the car's tires are $0.300 \mathrm{~m}$, what is the bump's period of oscillation? (c) What If? You hang a spring with spring constant $k=100 \mathrm{~N} / \mathrm{m}$ from the rear view mirror of your car. What is the mass that needs to be hung from this spring to produce simple harmonic motion with the same period as the bump on the tire? (d) What would be the maximum speed of the hanging mass in your car if you initially pulled the mass down $8.00 \mathrm{~cm}$ beyond equilibrium before releasing it?

While driving behind a car traveling at $3.00 \mathrm{~m} / \mathrm{s},$ you notice that one of the car's tires has a small hemispherical bump on its rim as shown in Figure $\mathrm{P} 15.16$.
(a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radii of the car's tires are $0.300 \mathrm{~m}$, what is the bump's period of oscillation? (c) What If? You hang a spring with spring constant $k=100 \mathrm{~N} / \mathrm{m}$ from the rear view mirror of your car. What is the mass that needs to be hung from this spring to produce simple harmonic motion with the same period as the bump on the tire? (d) What would be the maximum speed of the hanging mass in your car if you initially pulled the mass down $8.00 \mathrm{~cm}$ beyond equilibrium before releasing it?

Key Concepts

Simple Harmonic Motion (SHM)

SHM is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and acts in the direction opposite to the displacement. This results in a sinusoidal oscillation about an equilibrium point and can describe systems such as oscillating springs and the projection of uniform circular motion onto a line.

Projection of Uniform Circular Motion

This concept explains how the motion of a point moving at a constant speed around a circular path, when viewed from a fixed line (or axis), produces oscillatory, sinusoidal motion. It is relevant in scenarios where an object on a rotating body, such as a bump on a tire, appears to move back and forth in a simple harmonic pattern from a fixed viewpoint.

Period and Frequency of Oscillations

The period is the time taken for one complete cycle of oscillation, and frequency is the number of cycles per unit time. In oscillatory systems, these parameters relate to other quantities like the angular velocity in circular motion or the mass and spring constant in a spring-mass system, establishing a fundamental connection between the dynamical properties of the system.

Mass-Spring System and Hooke’s Law

A mass attached to a spring exhibits simple harmonic motion, which is governed by Hooke’s Law. The period of such a system is given by T = 2??(m/k), where m is the mass and k is the spring constant. This relationship is key for designing systems with specific oscillatory characteristics and is used to compare mechanical motions in different contexts.

Transcript

00:01 This problem, you are moving out of velocity of 3 meters per second in your car.

00:05 When you notice that there is this little bump on one of the wheels on the car in front of you.

00:14 And now in part a, what we're asked to do is explain why this looks like simple harmonic motion.

00:22 And so let's think about this.

00:24 And so what we're going to see here is basically this thing.

00:27 It's going to start out here.

00:28 And then go down like so, right? and then it's going to get here, and it's just going to oscillate back and forth.

00:36 What we're saying here is that this is going to look like a sign function.

00:41 And so let's kind of see why that is.

00:44 Let me do a side perspective here.

00:47 All right, so here's our bump.

00:49 This thing is just rotating at a constant rate, round and round, right? basically what i'm saying, is that the rate of change of your angle, d theta dt, if you will, or like omega, your rotational velocity is a constant, since you're moving at a constant speed.

01:14 And so this thing is constantly moving around, but we're not seeing this motion.

01:19 What we're seeing is this motion projected onto an axis like this, like a vertical axis.

01:27 And so what we can do here is we can define an angle like so, right? we can define a theta.

01:35 And of course the d -theta d -t here is going to be a constant.

01:39 But what we're interested in is what is the rejection of essentially this vector that points to this onto this vertical axis.

01:51 Well, what is that going to be? it's going to be, we'll call that distance, right, this height where it is like x, that x is going to be the radius of the wheel, whatever that is, times cosine of theta.

02:12 Now, cosine of theta, as i mentioned over here, since omega or the rate of change of our theta is constant, this is just going to be cosine of omega -t.

02:25 And now we see why it looks like harmonic motion, right? it's because you get this cosine in there.

02:32 And so, yeah, that's the reason why, is that this is going to have this cosine term, and this is going to change like a cosine function.

02:44 Okay, now, part b, we're asked to find what is the period of the oscillation, given that the radius is equal to 0 .3 meters.

02:56 Now here, we're going to use the fact that our velocity is equal to three, or three meters per second.

03:04 And so what this tells us, is that a rotational velocity or let's do it this way...

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