Problem 75 - Physics

Question

Interactive LearningWare 10.1 at reviews the concepts involved in this problem. 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 \(f=3.0\) \(\mathrm{Hz} ?\)

Step-by-Step Solution

Answer

Approximately 4.28 kg of mass should be attached to the spring.

Step 1: Understanding Hooke's Law

Hooke's Law relates the force exerted by a spring to its extension. It states that \( F = kx \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the extension of the spring. Here, we know the extension \( x = 0.018 \; \text{m} \) when a mass \( m = 2.8 \; \text{kg} \) is attached. The force due to the mass is the weight, \( F = mg \), where \( g \approx 9.8 \; \text{m/s}^2 \). This will help us find \( k \).

Step 2: Calculate the Spring Constant

Using the known values in Hooke's Law, substitute \( F = mg = 2.8 \times 9.8 \) and \( x = 0.018 \). Solve for \( k \): \[ k = \frac{F}{x} = \frac{2.8 \times 9.8}{0.018} = 1522.22 \; \text{N/m}.\]

Step 3: Relate Frequency and Mass

The frequency of vibration \( f \) of a spring-mass system is related to the spring constant \( k \) and the mass \( m \) as:\[f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}.\]We need to find the mass \( m \) for a frequency \( f = 3.0 \; \text{Hz}. \) Substitute \( k = 1522.22 \; \text{N/m} \) into the equation.

Step 4: Solve for the Desired Mass

Rearrange the frequency equation to solve for \( m \):\[ m = \frac{k}{(2\pi f)^2}.\]Substitute \( k = 1522.22 \; \text{N/m} \), and \( f = 3.0 \; \text{Hz} \):\[ m = \frac{1522.22}{(2\pi \times 3.0)^2} = \frac{1522.22}{(18.85)^2} = \frac{1522.22}{355.31} \approx 4.28 \; \text{kg}.\]

Step 5: Conclusion

The desired mass that needs to be attached to the spring to achieve a frequency of 3.0 Hz is approximately 4.28 kg.

Key Concepts

Spring Constant

The spring constant, often denoted by \( k \), plays a vital role in understanding how a spring responds to forces. It is a measure of the stiffness of the spring, indicating how much force is needed to extend or compress the spring by a certain distance.

When an object is attached to a spring and causes it to stretch, this stretch is described as its extension \( x \). According to Hooke's Law, the force \( F \) needed to extend or compress a spring by a distance \( x \) is proportional to \( x \) through the spring constant \( k \).

Thus, Hooke's Law is expressed as:

• \( F = kx \)

Here, \( F \) can also be represented by the weight \( mg \) of the object attached, where \( m \) is the mass and \( g \) is the acceleration due to gravity. From this relationship, you can solve for \( k \) if \( F \) and \( x \) are known, which helps understand the necessary conditions for specific applications, such as determining how much mass should be used to achieve a desired frequency of vibration.

Frequency of Vibration

The frequency of vibration of a mass-spring system reveals insights into how often the system oscillates. Understanding this is crucial for designing systems that need to exhibit specific oscillatory behavior.

The frequency \( f \) of a vibrating mass-spring system is determined by the formula:

• \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)

This equation shows that frequency depends on both the spring constant \( k \) and the mass \( m \) attached to the spring. This means that, given a more rigid spring or a lighter mass, the frequency increases.

If you know the desired frequency and spring constant, you can find the necessary mass to attach. This knowledge helps in numerous applications such as musical instruments, mechanical devices, and vibration control systems.

Mass-Spring System

The mass-spring system is a classic model in physics that illustrates how masses and springs interact through dynamic motion. This system is often used to study simple harmonic motion, where the restoring force is directly proportional to the displacement.

In such a system, the mass \( m \) hangs from a spring with a spring constant \( k \), leading to oscillations or vibrations when displaced from its equilibrium position. When discussing such systems:

• The displacement from the equilibrium involves potential and kinetic energy interchange.
• The period of oscillation and frequency are key characteristics that define the motion.

Mathematically, the period \( T \) of a mass-spring system is related to the frequency \( f \) as \( T = \frac{1}{f} \). These principles are critical for applications in engineering, physics education, and technology design, enabling control and prediction of system behavior.