A 3-kg mass is attached to a spring with stiffness k=48 N / m. The mass is displaced 1 / 2 m to

A 3-kg mass is attached to a spring with stiffness k=48 N /...

A 3-kg mass is attached to a spring with stiffness $$k=48 \mathrm{N} / \mathrm{m}$$. The mass is displaced $$1 / 2$$ m to the left of the equilibrium point and given a velocity of 2 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position?

Key Concepts

Initial Conditions and Equation of Motion

Initial conditions, such as initial displacement and initial velocity, are used to determine the specific constants in the general solution of the equation of motion for a mass-spring system. They allow one to tailor the general solution of the differential equation to reflect the unique behavior of the system at time zero, resulting in a complete description of the motion.

Amplitude, Period, and Frequency

The amplitude is the maximum displacement from the equilibrium position, the period is the time taken for one complete cycle of oscillation, and the frequency is the number of cycles per unit time. These quantities are interrelated and are determined by the energy of the system and the angular frequency. They provide critical insights into the quantitative behavior of oscillatory motion.

Simple Harmonic Motion

Simple harmonic motion (SHM) describes the type of oscillatory motion observed in systems where the restoring force is directly proportional to displacement and acts in the opposite direction. This motion is characterized by sinusoidal oscillations and occurs without energy loss when there is no damping. It serves as a foundational concept in mechanics and differential equations for modeling periodic behaviors.

Mass-Spring System

A mass-spring system is a classical mechanical model consisting of a mass attached to a spring. When displaced from its equilibrium position, the spring exerts a restoring force proportional to the displacement, leading to oscillatory motion. This model is often used to illustrate the principles of simple harmonic motion and solve second order linear differential equations.

Angular Frequency

Angular frequency, denoted by the Greek letter ?, quantifies the rate of oscillation in simple harmonic motion. It is calculated using the relation ? = ?(k/m), where k is the spring constant and m is the mass. Angular frequency is essential for determining the oscillatory behavior and appears in the standard form of the solution to the motion's differential equation.

Transcript

00:01 In this video, we're going to look at a spring that has a 3 kilogram mass attached and it has a 48 newton per meter stiffness.

00:10 So k is 48 newton per meter and the mass is 3 kilograms.

00:18 What this allows us to do is calculate the angular frequency omega, which is the square root of k over m, which is the square root of 48 over 3, square of 16, which is 4.

00:35 And we also know that initially it's displaced to the left, half a meter.

00:42 And so we know that the original position at times zero is minus half, minus because it's to the left.

00:48 And the initial velocity at time zero is two meters per second to the right.

00:53 And so this is positive two.

00:55 Okay.

00:56 Now we know that the differential equation is given by 3 y double prime or 3 y double prime plus 48 y is equal to 0.

01:14 Okay? there's no damping here.

01:18 And so what we have here is based on the solution of this type of equation, the general solution is given by some constant c1.

01:30 Times cosine of 5t because omega is 5, i'm sorry, 4, 4t, plus some other constant times sine of 4t.

01:54 Okay.

01:56 So if we use these initial conditions, then we can find c1 and c2.

02:04 So let's go ahead and do that.

02:06 If we plug in 0 into this equation, we get c1.

02:15 Cosine 0 is 1 plus 0 because sign 0 is 0 is equal to negative half, which means c1 is equal to negative half.

02:29 And if we plug in into the y prime, so y prime of t is 4 c1 times negative sign of 4 t plus 4c2 cosine, of 4t.

02:47 Okay.

02:48 So now if we plug in 0, we get cosine 0 is 1...

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