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The scale of a spring balance that reads from 0 to \(15.0 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz}\). (a) What is the spring constant? (b) How much does the package weigh?

Short Answer

Expert verified
The spring constant is approximately \(1226.25 \, \text{N/m}\). The package weighs approximately \( 30.4 \, \text{N} \) (or \( 3.1 \, \text{kg} \)).

Step by step solution

01

Understanding Hooke's Law and Spring Constant

The spring constant \( k \) is a measure of a spring's stiffness and is defined by Hooke's Law. When a spring is stretched or compressed by a distance \( x \), the force \( F \) exerted by the spring is proportional to the distance: \( F = kx \). The spring constant can be calculated if we know the maximum force the spring exerts when fully loaded, which corresponds to the weight it can handle, and the maximum stretch, which corresponds to the length of the scale.
02

Determine Maximum Stretch of Spring

The scale of the spring balance measures up to \(15.0 \, \mathrm{kg}\), which translates to the maximum force \( F_{max} = 15.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \) (the gravitational force). The length of the scale is given as \(12.0 \, \mathrm{cm} \), or \(0.12 \, \mathrm{m}\).
03

Calculate Spring Constant

Using Hooke’s Law, \( k = \frac{F}{x} \), where \( F = 15.0 \, \mathrm{kg} \times 9.81 \, \text{m/s}^2 = 147.15 \, \text{N} \) and \( x = 0.12 \, \mathrm{m} \). Thus, \( k = \frac{147.15}{0.12} \approx 1226.25 \, \text{N/m}\).
04

Relating Frequency to the Spring Constant

The frequency of oscillation \( f \) of a mass-spring system is given by \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \( m \) is the mass of the package. We know \( f = 2.00 \, \mathrm{Hz} \) and need to find \( m \).
05

Solve for Mass of the Package

Rearrange the frequency formula to solve for \( m \): \( m = \frac{k}{(2\pi f)^2} \). Substitute \( k = 1226.25 \, \text{N/m} \) and \( f = 2.00 \, \mathrm{Hz} \): \( m = \frac{1226.25}{(2\pi (2))^2} \approx 3.1 \, \mathrm{kg} \).
06

Calculate the Force Exerted by the Package

The weight of the package is given by the force exerted due to gravity \( W = mg \), where \( m \approx 3.1 \, \mathrm{kg} \) and \( g = 9.81 \, \text{m/s}^2 \). Thus, \( W \approx 3.1 \, \mathrm{kg} \times 9.81 \, \text{m/s}^2 \approx 30.4 \, \text{N} \).

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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 is a fundamental principle that defines the relationship between the force exerted by a spring and its displacement. When you apply a load to a spring, it stretches or compresses depending on the direction of the force.
The mathematical representation of Hooke's Law is:
  • \( F = kx \)
Here, \( F \) is the force exerted by the spring, \( k \) is the spring constant (which indicates how stiff the spring is), and \( x \) is the displacement of the spring from its original position. The spring constant \( k \) is unique for each spring and tells you how much force is required to move the spring a certain distance. A higher \( k \) means a stiffer spring.
In our exercise, the spring balance can handle a maximum load of \(15.0 \) kg, which translates to a gravitational force \( F_{max} = 147.15 \, \text{N}\). The spring stretches by \(0.12 \, \text{m}\), giving a spring constant \( k \approx 1226.25 \, \text{N/m}\). This constant represents the spring's resistance to being stretched or compressed.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium. It's the same kind of motion that a mass attached to a spring undergoes as it moves back and forth about a central point.
In the realm of springs, when a mass is attached, the system oscillates in a regular pattern if displaced. Important quantities in these systems include frequency \( f \) and period \( T \). In our exercise, the frequency is given as \(2.00 \, \text{Hz} \), meaning two complete oscillations occur each second.
The frequency of oscillation for a mass-spring system is defined by:
  • \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)
This formula shows that the frequency depends on the spring constant \( k \) and mass \( m \) attached to the spring. If you know the frequency and spring constant, you can determine the mass. For our scenario, rearranging the formula gives us an approximate mass of \(3.1 \, \text{kg} \). Simple Harmonic Motion is vital in understanding oscillatory systems in physics and engineering.
Mass-Spring System
A Mass-Spring System is a classic model in physics that consists of a mass attached to a spring. This system is used to study oscillatory motion and principles of dynamics.
In our example, the Mass-Spring System consists of a package tied to a spring balance, oscillating due to gravity. The spring constant \( k \) is vital in such systems as it determines the behavior of the oscillation. The system's frequency (2.00 Hz in our case) aids in calculating other components like the mass.
The equation for the frequency as mentioned earlier is
  • \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)
By knowing useful values like the spring constant and frequency, you can re-engineer to find missing quantities such as mass. In our exercise, after determining \( m \) as \(3.1 \, \text{kg} \), we can calculate the system’s weight, which is equivalent to the force due to gravity, \( W = mg \). This results in approximately \(30.4 \, \text{N}\). Mass-Spring Systems form the bedrock for understanding more complex oscillations and are extensively used to model real-world applications.

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Most popular questions from this chapter

For a simple pendulum, find the angular amplitude \(\theta_{m}\) at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by \(1.0 \%\). (See "Trigonometric Expansions" in Appendix E.)

A (hypothetical) large slingshot is stretched \(2.30 \mathrm{~m}\) to launch a \(170 \mathrm{~g}\) projectile with speed sufficient to escape from Earth (11.2 \(\mathrm{km} / \mathrm{s}\) ). Assume the elastic bands of the slingshot obey Hooke's law. (a) What is the spring constant of the device if all the elastic potential energy is converted to kinetic energy? (b) Assume that an average person can exert a force of \(490 \mathrm{~N}\). How many people are required to stretch the elastic bands?

A \(10 \mathrm{~g}\) particle undergoes SHM with an amplitude of \(2.0 \mathrm{~mm}\), a maximum acceleration of magnitude \(8.0 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\), and an unknown phase constant \(\phi\). What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00\) \(\mathrm{H} z\) (a) What is the spring constant of each spring if the mass of the car is \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car with an even distribution of mass?

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

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