Problem 11 - Advanced Engineering Mathematics

Question

A mass weighing 64 pounds stretches a spring \(0.32\) foot. The mass is initially released from a point 8 inches above the equilibrium position with a downward velocity of \(5 \mathrm{ft} / \mathrm{s}\). (a) Find the equation of motion. (b) What are the amplitude and period of motion? (c) How many complete cycles will the mass have completed at the end of \(3 \pi\) seconds? (d) At what time does the mass pass through the equilibrium position beading downward for the second time? (e) At what time does the mass attain its extreme displacement on either side of the equilibrium position? (f) What is the position of the mass at \(t=3\) s? (g) What is the instantaneous velocity at \(t=3 \mathrm{~s}\) ? (h) What is the acceleration at \(t=3 \mathrm{~s} ?\) (i) What is the instantaneous velocity at the times when the mass passes through the equilibrium position? (j) At what times is the mass 5 inches below the equilibrium position? (k) At what times is the mass 5 inches below the equilibrium position heading in the upward direction?

Step-by-Step Solution

Answer

The equation of motion is \( x(t) = -\frac{2}{3}\cos(10t) + \frac{1}{2}\sin(10t) \); amplitude = \( \frac{5}{6} \) ft, period = \( \frac{\pi}{5} \) s. Complete cycles in \( 3\pi \)= 15; equilibrium at \( \frac{\pi}{10} + \frac{1}{10}\arctan\frac{2}{3} \); extreme displacement \( \sim n\pi + \arctan\frac{1}{2} \); position at \( t=3 \) is \( -0.6415 \) ft; velocity \( t=3 \) is \( -4.16 \) ft/s; acceleration \( t=3 \) is \( 64.15 \) ft/s².

Step 1: Understanding the Problem

We need to model a mass-spring system using Hooke's Law and Newton's Second Law, then analyze its motion. The spring system has a mass of 64 pounds, stretches 0.32 feet, and is released from 8 inches above equilibrium with a downward velocity of 5 ft/s.

Step 2: Finding the Spring Constant

Use Hooke's Law, which relates force, spring constant, and displacement: \[ F = kx \] where force \( F = 64 \) pounds and displacement \( x = 0.32 \) feet. Thus, the spring constant \( k \) is calculated as: \[ k = \frac{F}{x} = \frac{64}{0.32} = 200 \text{ lb/ft} \]

Step 3: Writing the Differential Equation

The equation of motion for a spring-mass system (undamped) is given by: \[ m\frac{d^2x}{dt^2} + kx = 0 \] The mass \( m \) of the object is \( \frac{64}{32} = 2 \) slugs (since weight, \( w = mg \), \( 64 = mg \), \( g = 32 \text{ ft/s}^2 \)). The differential equation becomes: \[ 2\frac{d^2x}{dt^2} + 200x = 0 \] or, simplified: \[ \frac{d^2x}{dt^2} + 100x = 0 \]

Step 4: Solving the Differential Equation

The general solution of the differential equation \( \frac{d^2x}{dt^2} + 100x = 0 \) is: \[ x(t) = A\cos(10t) + B\sin(10t) \] where \( \omega = \sqrt{100} = 10 \). We need initial conditions to find \( A \) and \( B \).

Step 5: Applying Initial Conditions

The mass is released from 8 inches above equilibrium with a downward velocity of 5 ft/s (downward). Convert 8 inches to feet: \( x(0) = -\frac{8}{12} = -\frac{2}{3} \). The initial velocity gives \( x'(0) = 5 \). Solving the equations: 1. \( x(0) = A = -\frac{2}{3} \)2. \( v(t) = -10A\sin(10t) + 10B\cos(10t), \ v(0) = 10B = 5 \Rightarrow B = \frac{1}{2} \)Thus, the equation of motion is: \[ x(t) = -\frac{2}{3}\cos(10t) + \frac{1}{2}\sin(10t) \]

Step 6: Finding Amplitude and Period

The amplitude is calculated using:\[ A = \sqrt{\left(-\frac{2}{3}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{4}} = \sqrt{\frac{25}{36}} = \frac{5}{6} \text{ ft} \]The period \( T \) is given by: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ seconds} \]

Step 7: Calculating Complete Cycles After 3π Seconds

The number of cycles completed in \(3\pi\) seconds is:\[ \frac{\text{Total time}}{\text{Period}} = \frac{3\pi}{\frac{\pi}{5}} = 15 \text{ cycles} \]

Step 8: Finding Second Equilibrium Pass

Equilibrium crossings occur when \( x(t) = 0 \): \[ -\frac{2}{3}\cos(10t) + \frac{1}{2}\sin(10t) = 0 \Rightarrow \tan(10t) = \frac{2}{3} \]The first positive solution for \( t \) occurs at \( t = \frac{1}{10}\arctan\left(\frac{2}{3}\right) \). The second downward pass is \( t = \frac{\pi}{10} + \frac{1}{10}\arctan\left(\frac{2}{3}\right) \).

Step 9: Finding Extreme Displacements

Extremes occur when \( v(t) = 0 \), leading to:\[ -\frac{10}{3}\sin(10t) + 5\cos(10t) = 0 \Rightarrow \tan(10t) = \frac{5}{10} = \frac{1}{2} \]The solutions are symmetric and can be found by solving:\[ 10t = n\pi + \arctan\left(\frac{1}{2}\right) \] where \( n \) is an integer.

Step 10: Position at t=3s

Substitute \( t=3 \) into the equation of motion:\[ x(3) = -\frac{2}{3}\cos(30) + \frac{1}{2}\sin(30) \approx -0.667\cdot 0.154 + 0.5\cdot (-0.988) \approx -0.1475 - 0.494 \approx -0.6415 \text{ ft} \]

Step 11: Instantaneous Velocity at t=3s

Differentiate the equation of motion for velocity:\[ v(t) = 10\left(\frac{2}{3}\sin(10t) + \frac{1}{2}\cos(10t)\right) \]Substitute \( t=3 \):\[ v(3) = 10\left( \frac{2}{3}\sin(30) + \frac{1}{2}\cos(30) \right) \approx 10\left(-0.493 + 0.077 \right) \approx -4.16 \text{ ft/s} \]

Step 12: Acceleration at t=3s

Differentiate the velocity for acceleration:\[ a(t) = -100\left(-\frac{2}{3}\cos(10t) + \frac{1}{2}\sin(10t)\right) \]Substitute \( t=3 \):\[ a(3) = -100\left(-\frac{2}{3}\cdot0.154 + \frac{1}{2}\cdot-0.988\right) \approx -100\cdot(-0.1475 - 0.494) \approx 64.15 \text{ ft/s}^2 \]

Step 13: Instantaneous Velocity at the Equilibrium Position

The velocity when passing through the equilibrium, \( x=0 \), is calculated from:\[ v(t) = -10\times \left(-\frac{2}{3}\cos(10t) + \frac{1}{2}\sin(10t)\right) \Rightarrow v(t) = \frac{1}{2} \text{ ft/s downward} \]

Step 14: Times When Mass is 5 inches Below Equilibrium

Solve for \( x(t) = -\frac{5}{12} \): \[-\frac{2}{3}\cos(10t) + \frac{1}{2}\sin(10t) = -\frac{5}{12} \]

Step 15: Times When Mass is 5 inches Below Heading Up

To find the time when it heads upwards, solve \( -\frac{5}{12} \) and check when the velocity changes sign from negative to positive at that position.

Key Concepts

Hooke's Law

Hooke's Law is a cornerstone concept when studying mass-spring systems. It describes how the force needed to extend or compress a spring by a certain distance is proportional to that distance. This can be expressed with the formula:

• \( F = kx \)

Here, \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. In the context of the given problem, you first identify the spring constant by noting how much force (64 pounds) causes a specific stretch (0.32 feet).
The spring constant \( k \) is calculated by rearranging the formula to \( k = \frac{F}{x} \). For our sample problem, \( k = \frac{64}{0.32} = 200 \) lb/ft. This constant tells us how stiff the spring is, a higher value indicating a stiffer spring.

Newton's Second Law

Newton's Second Law forms the basis for describing the motion of a mass attached to a spring. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

• \( F = ma \)

For a mass-spring system, this is combined with Hooke's Law to form the equation of motion.
The weight provided in the exercise (64 pounds) helps determine the mass in slugs, because weight equals mass times the acceleration due to gravity \( g = 32 \text{ ft/s}^2 \). By dividing the weight by gravity, the mass \( m \) in slugs is \( \frac{64}{32} = 2 \).
You then set this against Hooke's Law-derived force to get a differential equation:

• \( 2\frac{d^2x}{dt^2} + 200x = 0 \)

This equation is pivotal for modeling the system's motion.

Differential Equations

Differential equations play a crucial role in modeling the behavior of mass-spring systems, particularly in the context of undamped motion. They are equations involving derivatives, expressing a relationship with the change rates of variables.

• The differential equation for an undamped mass-spring system:\[ \frac{d^2x}{dt^2} + 100x = 0 \]

In our scenario, the spring constant \( k \) and the mass \( m \) come together in this equation. Solving it helps you find \( x(t) \), describing how the position changes over time.
This type of second-order differential equation has a characteristic solution involving trigonometric functions:

• General Solution: \( x(t) = A\cos(\omega t) + B\sin(\omega t) \)

Where \( \omega \) is the angular frequency, derived from \( \omega = \sqrt{k/m} \), giving insights into the natural frequency of oscillation.

Undamped Motion

Undamped motion refers to the ideal scenario where a mass-spring system oscillates without energy loss, meaning no external forces like friction slow down the system. It's an important model in understanding basic oscillatory motion.

• In theory, the mass alternates between positions continuously over time, determined by its initial conditions: initial displacement and velocity.
• For our problem, these initial conditions help solve the specific solution for the general formula \( x(t) = A\cos(10t) + B\sin(10t) \).

The amplitude, which indicates the system's maximum displacement from equilibrium, and the period, which represents the complete cycle duration, are key characteristics of undamped motion.
This idealized model provides foundational insights that can be complicated by introducing damping factors like air resistance in other types of motion. But for undamped motion, energy remains constant.