Questions 1, 2, and 3 are short free-response questions that...

Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part. (DIAGRAM IS NOT AVAILABLE TO COPY) A horizontal spring with a force constant of 40 $\mathrm{N} / \mathrm{m}$ is attached a 0.1 $\mathrm{kg}$ block. (a) If the block is pulled to a distance of 0.5 m and released, what is the maximum speed of the block? (b) What is the frequency of the oscillations? (c) If the spring were flipped vertically and attached to the ground with the block placed on top, how would the natural length of the spring change? (d) How does the frequency of the oscillations of the vertical spring block oscillator compare with that when it was placed horizontally?

Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.
(DIAGRAM IS NOT AVAILABLE TO COPY)
A horizontal spring with a force constant of 40 $\mathrm{N} / \mathrm{m}$ is attached a 0.1 $\mathrm{kg}$ block.
(a) If the block is pulled to a distance of 0.5 m and released, what is the maximum speed of the block?
(b) What is the frequency of the oscillations?
(c) If the spring were flipped vertically and attached to the ground with the block placed on top, how would the natural length of the spring change?
(d) How does the frequency of the oscillations of the vertical spring block oscillator compare with that when it was placed horizontally?

Key Concepts

Spring Constant and Hooke's Law

The spring constant, k, quantifies the stiffness of a spring, as defined by Hooke's Law, which states that the force exerted by the spring is proportional to the displacement from its equilibrium position. This concept is fundamental in analyzing both energy storage in the spring and the dynamics of oscillations.

Oscillation Frequency and Angular Frequency

The oscillation frequency of a spring-mass system is determined by the mass and the spring constant, typically expressed by the formula ? = ?(k/m) for angular frequency, with the frequency given by f = ?/(2?). This relationship is key to understanding the time behavior of oscillatory motion.

Equilibrium and the Natural Length of a Spring

The natural length of a spring is its length when no external forces are acting on it. In vertical configurations, the weight of the attached mass causes a static extension, setting a new equilibrium position; however, this change in equilibrium does not alter the inherent properties of the spring or the system’s oscillatory frequency for small oscillations.

Simple Harmonic Motion (SHM)

SHM describes the oscillatory behavior of systems where the restoring force is proportional to the displacement from equilibrium, as per Hooke's Law. This concept allows us to analyze and predict characteristics such as maximum speed, frequency, and period in spring-mass systems.

Conservation of Mechanical Energy

In oscillatory systems without energy loss, mechanical energy is conserved. When a block attached to a spring is displaced, the maximum potential energy stored in the spring converts entirely into kinetic energy at equilibrium, enabling the calculation of maximum speed using energy equivalence.

Effect of Gravity in Vertical Oscillators

Gravity influences a vertically oriented spring-mass system by shifting the equilibrium position due to the weight of the mass. Despite this static displacement, the oscillatory motion characteristics—such as frequency—remain governed solely by the mass and the spring constant, assuming small amplitude oscillations.

Transcript

00:01 Here we're considering a simple spring mass system to look at some of the quantities that are involved when a mass starts oscillating on a spring.

00:11 So we're going to start off with a mass oscillating horizontally.

00:17 We are told its mass, like one kilogram, and its spring constant for the spring is 40 newtons per meter.

00:29 And what i'll suggest is we immediately know the frequency with which this mass will oscillate.

00:36 We'll pretend it's on a frictionless surface, so there is no friction.

00:41 But the angular frequency, omega, is simply equal to the square root of k over m.

00:51 So those two parameters govern how quickly it oscillates.

00:55 If we want that in terms of frequency in cycles per second, there's a simple relationship.

01:03 And if we use the numbers given 40 and 0 .1, what we get basically is 20 radians per second for the angular frequency.

01:22 And that means our frequency in cycles per second is just that divided by 2 pi.

01:30 It would be 10 over by cycles per second.

01:39 So roughly three cycles per second.

01:41 So we know already how this thing will oscillate once it gets set into motion.

01:47 But we can figure out another parameter.

01:50 Let's assume that we stretch the block immediately to the right by an amount of five centimeters.

02:02 It's a reasonable amount to stretch that spring.

02:07 What we can now calculate is not only the frequency of oscillation, but how fast it is going to oscillate back and forth.

02:16 We can simply use conservation of energy, that the original elastic energy that's in that system when you stretch it gets transferred all into kinetic energy when that mass goes through the equilibrium position.

02:38 So we'll consider that the point of highest velocity through the equilibrium position where the spring is neither stretched nor and of course kinetic energy is one half m v squared...

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