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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

You are given the task of opening an antiquated "light lock," which is unlocked by shining red light pulses of a certain frequency for a long duration of time into a light sensor on the lock. You are given a red laser pointer, a spring of unstretched length \(L=15.0 \mathrm{cm}\) and spring constant \(k=7.20 \mathrm{N} / \mathrm{m},\) a sheet of steel \(\left(\rho=7.60 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is 0.125 inches thick, and some tools. You come up with the idea to take a piece of the steel sheet (of mass \(m\) ), cut a slot in it, and hang it from the spring. If you shine the laser through the slot and onto the sensor, and then stretch the spring and let it go, the steel plate will oscillate and cause the beam to pass through the slot periodically. (a) Assuming the beam is passing through the slot (and onto the lock's sensor) when the spring-mass system is in equilibrium, how is the frequency at which the light pulses hit the sensor related to the frequency of the spring/mass (i.e., steel plate) system. (b) Based on your answer for (a), what should be the frequency of the spring/mass system if the unlocking frequency is \(3.50 \mathrm{Hz} ?\) (c) What should be the mass \(m\) of the steel plate? (d) Calculate some reasonable dimensions for the steel plate (i.e., they should be consistent with the mass that is required for the spring-mass system). You may assume the material removed to make the slot in the steel plate is of negligible mass.

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.)

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m},\) released from rest, and subsequently oscillate back and forth with an angular frequency of \(11.3 \mathrm{rad} / \mathrm{s} .\) What is the speed of the object at the instant when the spring is stretched by \(0.048 \mathrm{m}\) relative to its unstrained length?

A block of mass \(m=0.750 \mathrm{kg}\) is fastened to an unstrained horizontal spring whose spring constant is \(k=82.0 \mathrm{N} / \mathrm{m} .\) The block is given a displacement of \(+0.120 \mathrm{m},\) where the \(+\) sign indicates that the displacement is along the \(+x\) axis, and then released from rest. (a) What is the force (magnitude and direction) that the spring exerts on the block just before the block is released? (b) Find the angular frequency \(\omega\) of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

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?
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