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Magazine Chapter 10: Problem 21
A spring stretches by 0.018 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
The mass needed is approximately 4.293 kg.
Step by step solution
01
Understand the Problem
We need to find the mass that will cause a spring system to have a frequency of vibration equal to 3.0 Hz. We know that a 2.8 kg mass stretches the spring by 0.018 m. We can use this information to find the spring constant first.
02
Calculate the Spring Constant
Using Hooke's Law, we have the equation: \[ F = kx \]where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement (0.018 m). The force applied by a 2.8 kg mass is its weight: \[ F = mg = 2.8 \times 9.81 \approx 27.468 \text{ N} \]Therefore, the spring constant \( k \) can be calculated as:\[ k = \frac{F}{x} = \frac{27.468}{0.018} \approx 1526 \text{ N/m} \]
03
Use the Frequency Formula
The frequency \( f \) of a mass-spring system is given by:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]Rearranging for \( m \), we get:\[ m = \frac{k}{(2\pi f)^2} \]
04
Plug in the Known Values
Substitute \( k = 1526 \text{ N/m} \) and \( f = 3.0 \text{ Hz} \) into the formula:\[ m = \frac{1526}{(2\pi \times 3.0)^2} \]
05
Calculate the Mass
Compute \( (2\pi \times 3.0)^2 = (6\pi)^2 = 36\pi^2 \approx 355.3057 \). Then calculate:\[ m = \frac{1526}{355.3057} \approx 4.293 \text{ kg} \]
06
Verify and State the Result
The calculation shows that approximately 4.293 kg should be attached to the spring to achieve a 3.0 Hz frequency. Verify by plugging back into the equations if needed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hooke's Law
In the world of physics, Hooke's Law is a central concept when exploring spring motion. Hooke's Law describes the relationship between the force exerted on a spring and the spring鈥檚 displacement, mathematically expressed as \( F = kx \). Here, \( F \) represents the force applied to the spring, \( k \) denotes the spring constant, and \( x \) stands for the displacement from its equilibrium position.
Hooke's Law provides a fundamental understanding of how springs behave and reacts under different forces, serving as a stepping stone for further exploration like determining the spring constant.
- Linear Relationship: The law illustrates a linear relationship, meaning the force and displacement are directly proportional. This remains true as long as the deformation of the spring isn鈥檛 beyond its elastic limit.
- Essential for Calculations: Understanding this law is pivotal for calculating the spring constant, which is key to solving many physics problems.
Hooke's Law provides a fundamental understanding of how springs behave and reacts under different forces, serving as a stepping stone for further exploration like determining the spring constant.
Spring Constant Calculation
The spring constant \( k \) is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress the spring by a certain length. A higher spring constant means a stiffer spring.
Calculating the spring constant allows us to quantify the stiffness of the spring, which is crucial when predicting how the spring will behave under different masses or forces.
- Formula Use: Using the formula from Hooke's Law \( F = kx \), we can rearrange this to calculate \( k \) as \( k = \frac{F}{x} \).
- Example Calculation: In our exercise, a 2.8 kg mass applies a force of about 27.468 N (using \( F = mg \)), causing a displacement of 0.018 m. This leads to a spring constant calculation of \( k \approx 1526 \text{ N/m} \).
Calculating the spring constant allows us to quantify the stiffness of the spring, which is crucial when predicting how the spring will behave under different masses or forces.
Frequency of Vibration
The frequency of vibration for a mass-spring system is how often the mass oscillates back and forth per second. It depends on both the spring constant and the mass attached to the spring.
Understanding the frequency of vibration is essential for applications like tuning musical instruments or designing mechanisms where precise vibrations are crucial.
- Frequency Formula: The formula for frequency \( f \) is \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass.
- Calculating Mass: Rearranging this formula gives \( m = \frac{k}{(2\pi f)^2} \). This helps us find the mass needed for a spring system to achieve a desired frequency.
- Contextual Example: For instance, to achieve a frequency of 3.0 Hz, with \( k \approx 1526 \text{ N/m} \), the required mass is calculated to be approximately 4.293 kg.
Understanding the frequency of vibration is essential for applications like tuning musical instruments or designing mechanisms where precise vibrations are crucial.
Mass-Spring System Dynamics
A mass-spring system is a common model used to describe the dynamics of a mass attached to a spring. It effectively demonstrates how energy is conserved and transferred within a system.
In learning about mass-spring system dynamics, one gains a deeper appreciation for how mechanical systems operate and the fundamental physics principles that underlie these processes.
- Simple Harmonic Motion: Such systems often exhibit simple harmonic motion, oscillating back and forth due to the restoring force of the spring.
- Energy Transformation: The potential energy stored in the spring converts into kinetic energy and vice versa, illustrating the conservation of energy.
- Real-world Applications: Understanding this dynamic is vital for engineers and physicists when designing systems such as vehicle suspension, earthquake-resistant structures, and other mechanical devices.
In learning about mass-spring system dynamics, one gains a deeper appreciation for how mechanical systems operate and the fundamental physics principles that underlie these processes.
