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Chapter 5: Problem 1

A 12 -lb weight is placed upon the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring \(1.5\) in. The weight is then pulled down 2 in. below its equilibrium position and released from rest at \(t=0\). Find the displacement of the weight as a function of the time; determine the amplitude, period, and frequency of the resulting motion; and graph the displacement as a function of the time.

Short Answer

Expert verified
The displacement of the weight as a function of time is given by \(x(t) = 0.167\cos(32t)\), where t is in seconds. The amplitude of the oscillation is 0.167 ft, the period is approximately 0.196 s, and the frequency is about 5.102 Hz. The graph of this function will show a cosine wave with these characteristics.

Step by step solution

01

Determine the spring constant (k)

To determine the spring constant (k), we can use Hooke's Law, which states that the force required to stretch or compress a spring is proportional to the displacement of the spring. The formula for Hooke's Law is: \[F = -kx\] Where F is the force applied, k is the spring constant, and x is the displacement of the spring. In this case, the force is equal to the weight (12 lb) multiplied by the acceleration due to gravity (32 ft/s^2). The displacement of the spring is 1.5 in, which converts to 0.125 ft. Therefore, we can write the equation as: \[12 * 32 = k * 0.125\]
02

Solve for k

Now, we need to solve the equation to find the spring constant: \[k = \frac{12 * 32}{0.125}\] \[k = 3072\, \text{lb/ft}\]
03

Determine the angular frequency (蠅)

To find the angular frequency (蠅), we can use the formula: \[\omega = \sqrt{\frac{k}{m}}\] Where 蠅 is the angular frequency, k is the spring constant, and m is the mass. We can calculate the mass using the weight (12 lb) and the acceleration due to gravity (32 ft/s虏): \[m = \frac{12}{32} = 0.375\, \text{slug}\] Now, we can find the angular frequency: \[\omega = \sqrt{\frac{3072}{0.375}}\] \[\omega = 32\, \text{rad/s}\]
04

Write the equation for the displacement

The general equation for the displacement of a system undergoing simple harmonic motion is: \[x(t) = A\cos(\omega t + \varphi)\] Where A is the amplitude, 蠅 is the angular frequency, t is time, and 蠁 is the phase angle. Since the weight is released from 2 in (0.167 ft) below its equilibrium position, the amplitude A is 0.167 ft, and the phase angle 蠁 is 0. Therefore, the equation for the displacement as a function of time is: \[x(t) = 0.167\cos(32t)\]
05

Determine the amplitude, period, and frequency

The amplitude (A) is the maximum displacement of the system from its equilibrium position. In this case, the amplitude is 0.167 ft. The period (T) is the time it takes for the system to complete one full oscillation. To find the period, we can use the formula: \[T = \frac{2\pi}{\omega}\] \[T = \frac{2\pi}{32}\] \[T = 0.196\, \text{s}\] The frequency (f) is the number of oscillations per second. To find the frequency, we can use the formula: \[f = \frac{1}{T}\] \[f = \frac{1}{0.196}\] \[f = 5.102\, \text{Hz}\]
06

Graph the displacement as a function of time

Now, we can graph the displacement as a function of time using the equation: \[x(t) = 0.167\cos(32t)\] The graph should show a cosine wave with an amplitude of 0.167 ft, a period of 0.196 s, and a frequency of approximately 5.102 Hz.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hooke's Law
When we talk about **Hooke's Law**, we're discussing a principle that defines how certain materials (like springs) behave when forces are applied to them. It's quite simple: the force needed to either stretch or compress a spring is directly proportional to the distance it moves from its resting position, also known as displacement. Mathematically, this is expressed as:

\[F = -kx\] where:
  • \(F\) is the force applied to the spring,
  • \(k\) is the spring constant, which tells us how stiff the spring is,
  • \(x\) is the displacement from the spring's original position.
The negative sign indicates that the direction of force is opposite to the direction of displacement. In the example from the exercise, the spring stretches by 1.5 inches due to a 12-pound weight, illustrating how Hooke's Law can be used to find the spring constant easily.
Spring Constant
The **spring constant** is a crucial value that measures the stiffness of a spring. In essence, it tells us how much force is needed to stretch the spring by a unit length. With Hooke's Law in mind, we determine this constant when we know the force applied and the displacement it causes. The formula we use is:

\[k = \frac{F}{x}\]
  • In the previous example, the weight was exerting a force of \(12 \times 32\, \text{lb-ft/s}^2\).
  • The spring stretched 1.5 inches, which converts to 0.125 feet.
  • Solving for \(k\), we find it to be \(3072\, \text{lb/ft}\).
This specific value helps us in further calculations related to simple harmonic motion, like determining angular frequency.
Angular Frequency
**Angular frequency**, denoted by the Greek letter \(\omega\), describes how fast something oscillates through a cycle, particularly in circular motion or vibration contexts like with springs. It's calculated using the relationship:

\[\omega = \sqrt{\frac{k}{m}}\] where:
  • \(k\) is the spring constant,
  • \(m\) is the mass of the object attached to the spring.
For our specific problem, we established the spring constant \(k\) as 3072 lb/ft and calculated the mass \(m\) from the given weight is 0.375 slugs. Plugging these into the formula, we find the angular frequency as 32 rad/s. This tells us the rate at which the spring oscillates back and forth.
Displacement Equation
The **displacement equation** in simple harmonic motion illustrates how the position of an oscillating object varies over time. For a spring system, the general equation is:

\[x(t) = A\cos(\omega t + \varphi)\] where:
  • \(A\) is the amplitude, or the maximum displacement from the equilibrium position,
  • \(\omega\) is the angular frequency,
  • \(t\) represents time,
  • \(\varphi\) is the phase angle, which describes the starting point in the cycle.
In the exercise scenario, the displacement equation becomes \(x(t) = 0.167\cos(32t)\). Here, the amplitude \(A\) is 0.167 ft, indicating how far the object moves from its equilibrium, and the phase angle \(\varphi\) is zero, since it starts from its maximum displacement. This equation helps us visualize the oscillation pattern over time.

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Most popular questions from this chapter

A 4 -lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 in. At time \(t=0\) the weight is then struck so as to set it into motion with an initial velocity of \(2 \mathrm{ft} / \mathrm{sec}\), directed downward. (a) Determine the resulting displacement and velocity of the weight as functions of the time. (b) Find the amplitude, period, and frequency of the motion. (c) Determine the times at which the weight is \(1.5\) in. below its equilibrium position and moving downward. (d) Determine the times at which it is \(1.5\) in. below its equilibrium position and moving upward.

An 8-lb weight is attached to the lower end of a coil spring suspended from the ceiling and comes to rest in its equilibrium position, thereby stretching the spring \(0.4 \mathrm{ft}\). The weight is then pulled down 6 in. below its equilibrium position and released at \(t=0\). The resistance of the medium in pounds is numerically equal to \(2(d x / d t)\), where \(d x / d t\) is the instantaneous velocity in feet per second. (a) Set up the differential equation for the motion and list the initial conditions. (b) Solve the initial-value problem set up in part (a) to determine the displacement of the weight as a function of the time. (c) Express the solution found in step (b) in the alternative form \((5.32)\) of the text. (d) What is the so-called "period" of the motion? (e) Graph the displacement as a function of the time.

A circuit has in series an electromotive force given by \(E(t)=5 \sin 100 t \mathrm{~V}\), a resistor of \(10 \Omega\), an inductor of \(0.05 \mathrm{H}\), and a capacitor of \(2 \times 10^{-4}\) farads. If the initial current and the initial charge on the capacitor are both zero, find the charge on the capacitor at any time \(t>0\).

A 20-lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 in. Various external forces of the form \(F(t)=\) cos wt are applied to the system and it is found that the resonance frequency is \(0.5\) cycles/sec. Assuming that the resistance of the medium in pounds is numerically equal to \(a(d x / d t)\), where \(d x / d t\) is the instantaneous velocity in feet per second, determine the damping coefficient \(a\).

A 16 -lb weight is attached to the lower end of a coil spring that is suspended vertically from a support and for which the spring constant \(k\) is \(10 \mathrm{lb} / \mathrm{ft}\). The weight comes to rest in its equilibrium position and is then pulled down 6 in. below this position and released at \(t=0 .\) At this instant the support of the spring begins a vertical oscillation such that its distance from its initial position is given by \(\frac{1}{2} \sin 2 t\) for \(t \geq 0\). The resistance of the medium in pounds is numerically equal to \(2(d x / d t)\), where \(d x / d t\) is the instantaneous velocity of the moving weight in feet per second. (a) Show that the differential equation for the displacement of the weight from its equilibrium position is $$ \frac{1}{2} \frac{d^{2} x}{d t^{2}}=-10(x-y)-2 \frac{d x}{d t}, \quad \text { where } \quad y=\frac{1}{2} \sin 2 t $$ and hence that this differential equation may be written $$ \frac{d^{2} x}{d t^{2}}+4 \frac{d x}{d t}+20 x=10 \sin 2 t $$ (b) Solve the differential equation obtained in step (a), apply the relevant initial conditions, and thus obtain the displacement \(x\) as a function of time.
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