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Chapter 5: Problem 1

A 4-pound weight is attached to a spring whose spring constant is \(16 \mathrm{lb} / \mathrm{ft}\). What is the period of simple harmonic motion?

Short Answer

Expert verified
The period of simple harmonic motion is approximately 0.55 seconds.

Step by step solution

01

Identify the formula for period

For simple harmonic motion, the period \( T \) of a mass-spring system is given by the formula:\[ T = 2\pi \sqrt{\frac{m}{k}} \]where \( m \) is the mass of the object (in slugs) and \( k \) is the spring constant.
02

Convert the weight to mass

The weight of the object is given as 4 pounds. To find the mass \( m \) in slugs, use the fact that weight \( W = mg \), where \( g \approx 32.2 \ \mathrm{ft/s^2} \) is the acceleration due to gravity. Thus,\[ m = \frac{W}{g} = \frac{4}{32.2} \approx 0.124 \ \text{slugs} \]
03

Substitute values into the period formula

Now that we have \( m = 0.124 \ \text{slugs} \) and \( k = 16 \ \mathrm{lb/ft} \), substitute these into the period formula:\[ T = 2\pi \sqrt{\frac{0.124}{16}} \]
04

Calculate the period

Carry out the calculation:- Calculate the fraction: \( \frac{0.124}{16} = 0.00775 \)- Find the square root: \( \sqrt{0.00775} \approx 0.088 \)- Multiply by \( 2\pi \):\[ T \approx 2 \times 3.1416 \times 0.088 \approx 0.55 \ \mathrm{seconds} \]

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Key Concepts

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

Spring Constant
The spring constant, denoted as \( k \), plays a critical role in the dynamics of a spring in simple harmonic motion. It essentially quantifies the stiffness of the spring. A larger spring constant means a stiffer spring, while a smaller constant indicates a more flexible spring.
The spring constant is measured in \ \( \text{lb/ft} \) in the imperial system or \( \text{N/m} \) in the metric system.
In essence, the spring constant determines how much force is needed to stretch or compress the spring by a given amount. This property is captured in Hooke鈥檚 Law, which states:
  • \( F = -kx \)
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. This relationship is foundational for understanding the behavior of springs in motion. Understanding the spring constant will help you gauge the energy stored in a spring and predict how it will behave under various forces.
Mass-Spring System
A mass-spring system is a classic way to study simple harmonic motion, a type of periodic motion.
In this system, the mass is attached to a spring, and when displaced from its equilibrium position and released, it oscillates around that position. This setup forms the core principle behind many mechanical devices, such as watches and vehicular suspensions.
The mass-spring system follows Hooke's Law, implying that the force exerted by the spring is proportional to the distance it is stretched or compressed.
  • The key components of the mass-spring system are the mass \( m \) and spring constant \( k \). The interplay between these components determines how the system oscillates.
  • When displaced, the spring exerts a force that attempts to restore the mass to equilibrium.
As it moves, the mass exchanges potential energy stored in the spring with kinetic energy of motion. This harmonious exchange is what maintains the oscillation over time.
Period of Oscillation
The period of oscillation refers to the time it takes for the mass-spring system to complete one full cycle of motion.
Mathematically, the period \( T \) is crucial because it tells us how frequently the system oscillates. The period is determined by the formula:
  • \( T = 2\pi \sqrt{\frac{m}{k}} \)
This equation highlights that the period \( T \) depends on:
  • \( m \) 鈥� the mass of the object attached to the spring (in slugs)
  • \( k \) 鈥� the spring constant
From this formula, it's clear that a heavier mass or a less stiff spring (smaller \( k \)) will result in a longer period or slower oscillation. Conversely, a lighter mass or a stiffer spring will increase the frequency of oscillation. By manipulating \( m \) or \( k \), you can control the pace at which the system returns to its starting position after each cycle.

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Most popular questions from this chapter

\(y^{\prime \prime}+\lambda^{2} y=0, \quad y(0)=0, y(L)=0\)

The given differential equation is a model of an undamped spring/mass system in which the restoring force \(F(x)\) in (1) is nonlinear. For each equation use an ODE solver to obtain the solution curves satisfying the given initial conditions. If the solutions are periodic, use the solution curve to estimate the period \(T\) of oscillations. $$ \begin{aligned} &\frac{d^{2} x}{d t^{2}}+2 x-x^{2}=0 \\ &x(0)=1, x^{\prime}(0)=1 ; \quad x(0)=\frac{3}{2}, x^{\prime}(0)=-1 \end{aligned} $$

A 1-kilogram mass is attached to a spring whose constant is \(16 \mathrm{~N} / \mathrm{m}\), and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity. Determine the equations of motion if (a) the weight is released from rest 1 meter below the equilibrium position and (b) the weight is released 1 meter below the equilibrium position with an upward velocity of \(12 \mathrm{~m} / \mathrm{s}\).

A series circuit contains an inductance of \(L=1 \mathrm{~h}\), a capacitance of \(C=10^{-4} \mathrm{f}\), and an electromotive force of \(E(t)=100 \sin 50 t \mathrm{~V}\). Initially the charge \(q\) and current \(i\) are zero. (a) Find the equation for the charge at any time. (b) Find the equation for the current at any time. (c) Find the times for which the charge on the capacitor is zero.

Find the charge on the capacitor in an \(L R C\) series circuit at \(t=0.01 \mathrm{~s}\) when \(L=0.05 \mathrm{~h}, R=2 \Omega, C=0.01 \mathrm{f}, E(t)=0 \mathrm{~V}, q(0)=5 \mathrm{C}\), and \(i(0)=0 \mathrm{~A}\). Determine the first time at which the charge on the capacitor is equal to zero.
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