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Chapter 10: Problem 21

A spring stretches by \(0.018 \mathrm{m}\) when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0 \mathrm{Hz} ?\)

Short Answer

Expert verified
Attach approximately 2.70 kg for the spring to vibrate at 3.0 Hz.

Step by step solution

01

Understanding Hooke's Law and Frequency

First, we need to understand that the spring extends by a distance when a force is applied, according to Hooke's Law: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the extension of the spring. Additionally, the frequency of the spring-mass system is given by \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \( m \) is the mass attached to the spring.
02

Calculating the Spring Constant

To find the spring constant \( k \), we use Hooke's Law. The force due to the weight of the first mass is \( F = mg = 2.8 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 27.44 \, \text{N} \). Since \( F = kx \), we have \( k = \frac{F}{x} = \frac{27.44 \, \text{N}}{0.018 \, \text{m}} = 1524.44 \, \text{N/m} \).
03

Setting Up the Frequency Equation

We need the system to have a frequency of \( f = 3.0 \, \text{Hz} \). The frequency equation \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \) becomes \( 3.0 = \frac{1}{2\pi} \sqrt{\frac{1524.44}{m}} \).
04

Solving for Mass

To find \( m \), rearrange the equation: \( 3.0 = \frac{1}{2\pi} \sqrt{\frac{1524.44}{m}} \). Solving for \( m \):1. Multiply both sides by \( 2\pi \) to clear the fraction: \( 3.0 \times 2\pi = \sqrt{\frac{1524.44}{m}} \).2. Square both sides to remove the square root: \( (3.0 \times 2\pi)^2 = \frac{1524.44}{m} \).3. Rearrange to find \( m \): \( m = \frac{1524.44}{(3.0 \times 2\pi)^2} \).4. Calculate \( m \): \( m \approx 2.70 \, \text{kg} \).

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Key Concepts

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

Hooke's Law
Hooke's Law provides a fundamental principle that describes how springs behave under load. It鈥檚 a simple yet powerful idea. When a spring is either compressed or stretched, it exerts a force proportional to the displacement. The mathematical expression of Hooke's Law is given by:\[ F = kx \]where:
  • \( F \) is the force exerted by the spring, measured in Newtons (N).
  • \( k \) is the spring constant, representing the stiffness of the spring, measured in Newtons per meter (N/m).
  • \( x \) is the displacement or stretch/compression from the spring鈥檚 equilibrium position, measured in meters (m).
Understanding this principle means that if you double the force applied to the spring, the displacement will also double, assuming the spring doesn鈥檛 reach its elastic limit. This law is valid only when the stress and strain remain linear.
It鈥檚 essential for many phenomena, including vibrations in spring-mass systems, and it helps predict how springs will behave under various forces.
Spring Constant
The spring constant \( k \) is a crucial parameter in the behavior of springs. It quantifies the stiffness of a spring in Hooke's Law. The higher the spring constant, the stiffer the spring is.
In practical terms, a stiff spring with a high \( k \) value requires more force to produce the same amount of extension or compression than a spring with a lower \( k \).
To find the spring constant, you can rearrange Hooke's Law:\[ k = \frac{F}{x} \]This equation shows that the spring constant is the ratio of the force applied to the extension produced. In the original exercise, when a 2.8 kg object results in a force of 27.44 N, and stretches the spring by 0.018 m, we find that:\[ k = \frac{27.44}{0.018} = 1524.44 \, \text{N/m} \]Such a high spring constant indicates a very stiff spring. Knowing the spring constant is essential for predicting how the spring will react in various applications, such as in sensing devices, automotive suspensions, and vibration damping systems.
Frequency of Vibration
The frequency of vibration in a spring-mass system tells us how often the system oscillates back and forth. It is crucial for understanding how energy moves through the system.
In a spring-mass system, the frequency \( f \) depends on both the spring constant \( k \) and the mass \( m \) attached to it. The relation is expressed as:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]This formula shows:
  • As the spring constant increases, the frequency of vibration also increases. A stiffer spring vibrates faster.
  • Conversely, as the mass increases, the frequency decreases. A heavier mass slows down the vibration rate.
In the exercise, we calculated the necessary mass to achieve a vibration frequency of 3.0 Hz. Rearranging the frequency equation allowed us to find the mass that would keep the system vibrating at this specific frequency, giving us an approximate value of 2.70 kg.
Understanding this relationship is vital for numerous technological applications, from tuning musical instruments to designing stable structures and various mechanical systems.

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Most popular questions from this chapter

A \(0.60-\mathrm{kg}\) metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to \(0.23 \mathrm{m}\) (relative to its unstrained length), the speed of the sphere decreases from 5.70 to \(4.80 \mathrm{m} / \mathrm{s}\). What is the spring constant of the spring?

A spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg}\), and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m}\). The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74 .\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

A simple pendulum is made from a 0.65-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest speed?

A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 9.2 s. What is the height of the tower?

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to \(3.0 \times 10^{-2} \mathrm{N}\). The length and radius of the collagen are, respectively, 2.5 and \(0.091 \mathrm{cm},\) and Young's modulus is \(3.1 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}\). (a) If the stretching obeys Hooke's law, what is the spring constant \(k\) for collagen? (b) How much work is done by the variable force that stretches the collagen? (See Section 6.9 for a discussion of the work done by a variable force.)
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