Problem 91 - Calculus Volume 3

Question

A block has a mass of \(5 \mathrm{~kg}\) and is attached to a vertical spring with a spring constant of \(20 \mathrm{~N} / \mathrm{m}\). The block is released from the equilibrium position with a downward velocity of \(10 \mathrm{~m} / \mathrm{sec}\) a. Find the position function \(x(t)\) of the block. b. Find the period and frequency of the vibration. c. Sketch a graph of \(x(t)\). d. At what time does the block first pass through the equilibrium position?

Step-by-Step Solution

Answer

x(t) = 5 \sin(2t); Period: \(\pi\) s; First equilibrium pass at \(t = \frac{\pi}{2}\).

Step 1: Write the Differential Equation

The motion of the block is described by the simple harmonic motion differential equation given by \[ m \frac{d^2 x}{dt^2} + kx = 0,\]where \(m = 5 \; \mathrm{kg}\) is the mass and \(k = 20 \; \mathrm{N/m}\) is the spring constant.

Step 2: Determine the Angular Frequency

The angular frequency \(\omega\) for a mass-spring system is defined as \(\omega = \sqrt{\frac{k}{m}}\). Substituting the given values, we have: \[\omega = \sqrt{\frac{20}{5}} = \sqrt{4} = 2 \; \mathrm{rad/s}.\]

Step 3: Write the General Solution for the Position Function

The general solution for the position \(x(t)\) in a simple harmonic motion is \[x(t) = A \cos(\omega t) + B \sin(\omega t),\]where \(A\) and \(B\) are constants determined by initial conditions.

Step 4: Apply Initial Conditions

Given that the block is released from the equilibrium with a velocity, we have initial conditions:- At \(t = 0\), \(x(0) = 0\), implying \(A = 0\).- The initial downward velocity is \(v(0) = 10 \; \mathrm{m/s}\). Differentiate \(x(t)\) to find the velocity:\[v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t) + B \omega \cos(\omega t).\]Substitute the initial condition:\[v(0) = 2B = 10 \implies B = 5.\]Thus, the position function becomes \[x(t) = 5 \sin(2t).\]

Step 5: Find the Period

The period \(T\) of the vibration is given by \[T = \frac{2\pi}{\omega}.\]Substituting the value of \(\omega\), we have:\[T = \frac{2\pi}{2} = \pi \; \mathrm{seconds}.\]

Step 6: Calculate the Frequency

The frequency \(f\) is the reciprocal of the period:\[f = \frac{1}{T} = \frac{1}{\pi} \approx 0.318 \; \mathrm{Hz}.\]

Step 7: Sketch the Graph of the Position Function

The function \(x(t) = 5\sin(2t)\) represents a sinusoidal motion. It has an amplitude of 5, a period of \(\pi\) seconds, and oscillates symmetrically about the equilibrium position at 0. The graph will look like a regular sine wave, starting from the origin, moving upwards first.

Step 8: Determine First Equilibrium Pass Time

The equilibrium position is passed when \(x(t) = 0\). We solve for the first positive time when the block returns to equilibrium:\[\5 \sin(2t) = 0 \implies \sin(2t) = 0.\]This occurs initially when \(2t = \pi\), so:\[t = \frac{\pi}{2} \; \mathrm{seconds}.\]

Key Concepts

Differential Equation

The concept of a differential equation is central in understanding simple harmonic motion, especially in systems like a block attached to a spring. A differential equation is a mathematical equation that relates a function to its derivatives. In this context, we deal with a second-order linear differential equation, which describes the motion of the system. For a mass-spring system undergoing simple harmonic motion, the standard differential equation is given by: \[ m \frac{d^2 x}{dt^2} + kx = 0, \]where:- \(m\) is the mass of the object,- \(\frac{d^2 x}{dt^2}\) represents the acceleration,- \(k\) is the spring constant,- and \(x\) is the displacement from the equilibrium position.This equation highlights two opposing forces: the inertia of the mass and the restoring force of the spring. Solving this gives us insights into how the system evolves over time.

Angular Frequency

Angular frequency, denoted as \(\omega\), is a crucial concept in oscillatory motion. It helps us understand how fast an object oscillates back and forth within a mass-spring system. Mathematically, angular frequency is given by:\[ \omega = \sqrt{\frac{k}{m}}, \]where:- \(k\) is the spring constant,- \(m\) is the mass of the object.This formula shows that the angular frequency depends on the characteristics of the spring and the mass. A higher spring constant or a lighter mass will result in a higher angular frequency. In our exercise, the calculations yield \(\omega = 2 \; \text{rad/s}\), indicating each complete oscillation occurs relatively quickly.

Vibration Period

The vibration period \(T\) is the time required for one complete cycle of oscillation. It's a key property of simple harmonic motion. To find it, we use the angular frequency:\[ T = \frac{2\pi}{\omega}. \]This formula shows that the period is inversely proportional to the angular frequency. If the angular frequency increases, the period decreases, meaning the system oscillates faster. From our earlier calculation, with \(\omega = 2 \; \text{rad/s}\), the period \(T\) is calculated as \(\pi \; \text{seconds}\). This gives a precise timeframe for each full oscillation, helping us predict the system's behavior over time.

Mass-Spring System

A mass-spring system is a fundamental model for understanding simple harmonic motion. In this system, a mass is connected to a spring, which exerts a force directly proportional to the displacement of the mass, following Hooke's Law.The equation of motion derives from Newton's second law and Hooke's law, combined to form the differential equation:\[ m \frac{d^2 x}{dt^2} + kx = 0, \]where each parameter is crucial:- The mass \(m\) influences inertia.- The spring constant \(k\) determines the stiffness of the spring.- The displacement \(x\) varies as the system oscillates.This system provides a relatable example of oscillatory dynamics, seen in everyday life, from car suspensions to playground swings. It's a simplified model that captures the essence of oscillation, making it a powerful tool for learning about periodic motion.