A Weight Attached To A Spring Is At Its Lowest Point, 9 Inches Below Equilibrium, At Time $t=0$ Seconds. When The Weight Is Released, It Oscillates And Returns To Its Original Position At $t=3$ Seconds. Which Of The Following

Mar 12, 2025 by ADMIN 230 views

**A Weight Attached to a Spring: Understanding Oscillations**

Introduction

In this article, we will delve into the world of simple harmonic motion, where a weight attached to a spring oscillates between its equilibrium position and a lower point. We will analyze the given scenario, where the weight is initially at its lowest point, 9 inches below equilibrium, at time $t=0$ seconds, and returns to its original position at $t=3$ seconds. Our goal is to understand the underlying physics and determine the characteristics of this oscillation.

Understanding Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of periodic motion where an object oscillates about a fixed equilibrium position. The motion is characterized by a restoring force that is proportional to the displacement from the equilibrium position. In the case of a weight attached to a spring, the restoring force is provided by the spring's elasticity.

Mathematical Model

The mathematical model for SHM is given by the equation:

x(t) = A \cos(\omega t + \phi)

where $x(t)$ is the displacement from the equilibrium position at time $t$ , $A$ is the amplitude of the oscillation, $\omega$ is the angular frequency, and $\phi$ is the phase angle.

Given Scenario

In the given scenario, the weight is initially at its lowest point, 9 inches below equilibrium, at time $t=0$ seconds. This means that the displacement at $t=0$ is $x(0) = -9$ inches. The weight returns to its original position at $t=3$ seconds, which means that the displacement at $t=3$ is $x(3) = 0$ inches.

Determining the Characteristics of the Oscillation

To determine the characteristics of the oscillation, we need to find the amplitude, angular frequency, and phase angle. We can use the given information to set up a system of equations.

Equation 1: Displacement at $t=0$

x(0) = -9 = A \cos(\phi)

Equation 2: Displacement at $t=3$

x(3) = 0 = A \cos(3\omega + \phi)

Equation 3: Angular Frequency

We know that the weight returns to its original position at $t=3$ seconds, which means that the angular frequency is related to the period of the oscillation. The period $T$ is given by:

T = \frac{2\pi}{\omega}

Since the weight returns to its original position at $t=3$ seconds, the period is $T = 3$ seconds. Therefore, the angular frequency is:

\omega = \frac{2\pi}{T} = \frac{2\pi}{3}

Solving the System of Equations

We can now substitute the value of $\omega$ into Equation 2 and solve for $A$ and $\phi$ .

0 = A \cos\left(3\left(\frac{2\pi}{3}\right) + \phi\right)

0 = A \cos(2\pi + \phi)

Since $\cos(2\pi + \phi) = \cos(\phi)$ , we have:

0 = A \cos(\phi)

This implies that $A = 0$ or $\cos(\phi) = 0$ . However, since the weight is initially at its lowest point, we know that $A \neq 0$ . Therefore, we must have $\cos(\phi) = 0$ , which implies that $\phi = \frac{\pi}{2}$ .

Substituting the Value of $\phi$ into Equation 1

Now that we have found the value of $\phi$ , we can substitute it into Equation 1 to find the value of $A$ .

-9 = A \cos\left(\frac{\pi}{2}\right)

Since $\cos\left(\frac{\pi}{2}\right) = 0$ , we have:

-9 = 0

This is a contradiction, which means that our assumption that $\phi = \frac{\pi}{2}$ is incorrect. However, we can try another value of $\phi$ .

Alternative Solution

Let's try $\phi = \frac{3\pi}{2}$ . Substituting this value into Equation 1, we get:

-9 = A \cos\left(\frac{3\pi}{2}\right)

Since $\cos\left(\frac{3\pi}{2}\right) = 0$ , we have:

-9 = 0