Chapter 7: Problem 89
A 400-g mass stretches a spring \(5 \mathrm{~cm}\). Find the equation of motion of the mass if it is released from rest from a position \(15 \mathrm{~cm}\) below the equilibrium position. What is the frequency of this motion?
Short Answer
Expert verified
The equation of motion is \( y(t) = 0.15 \cos(14t) \), and the frequency is approximately 2.23 Hz.
Step by step solution
01
Understand the Problem
A mass-spring system is being described where a 400-gram mass stretches a spring by 5 cm. If this mass is released from a position 15 cm below the equilibrium, we need to determine the equation of motion and frequency of oscillation. Begin by identifying the forces and parameters involved: spring constant (k), mass (m), and initial conditions.
02
Calculate the Spring Constant (k)
Hooke's Law states that force is proportional to the displacement (F = kx). The weight of the mass (mg) causes the spring to stretch 5 cm at equilibrium. Using \( F = mg \), where \( m = 0.4 \) kg (400 g) and \( g = 9.8 \) m/s²:\[ k = \frac{mg}{x} = \frac{0.4 \times 9.8}{0.05} = 78.4 \, \text{N/m} \]
03
Determine Angular Frequency (ω)
The angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{k}{m}} \). Using \( k = 78.4 \, \text{N/m} \) and \( m = 0.4 \, \text{kg} \):\[ \omega = \sqrt{\frac{78.4}{0.4}} = \sqrt{196} = 14 \, \text{rad/s} \]
04
Formulate the Equation of Motion
For a mass-spring system released from rest at a displacement of 15 cm, the equation is given by: \[ y(t) = A \cos(\omega t + \phi) \]Where \( A = 0.15 \, \text{m} \) (converted from cm), and as it's released from rest, \( \phi = 0 \):\[ y(t) = 0.15 \cos(14t) \]
05
Calculate the Frequency
The frequency \( f \) is derived from angular frequency \( \omega = 2\pi f \):\[ f = \frac{\omega}{2\pi} = \frac{14}{2\pi} \approx 2.23 \, \text{Hz} \]
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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 concept in physics that deals with the behavior of springs and elastic materials. It states that the force required to extend or compress a spring by a certain distance is directly proportional to that distance. Mathematically, this relationship is expressed as:
In our example, when a 400-g mass causes the spring to stretch by 5 cm, the weight of the mass is balanced by the spring force at equilibrium. By using Hooke's Law and the weight of the mass, the spring constant \( k \) can be calculated. This concept is essential for understanding how springs behave in various physical systems.
- \( F = kx \)
In our example, when a 400-g mass causes the spring to stretch by 5 cm, the weight of the mass is balanced by the spring force at equilibrium. By using Hooke's Law and the weight of the mass, the spring constant \( k \) can be calculated. This concept is essential for understanding how springs behave in various physical systems.
angular frequency
Angular frequency is a key concept in understanding oscillatory motions, such as those of a mass-spring system. It is denoted by the symbol \( \omega \) and represents how fast an object oscillates in a circular path. Angular frequency is related to the spring constant and the mass of the object in the system through the equation:
- \( \omega = \sqrt{\frac{k}{m}} \)
equation of motion
The equation of motion describes how a physical system evolves over time. For a mass-spring system, this equation is typically sinusoidal due to the nature of simple harmonic motion. The general form of the equation of motion for such a system is:
In this equation:
Thus, the specific equation of motion in our scenario is \( y(t) = 0.15 \cos(14t) \), depicting how the mass moves over time from its initial position.
- \( y(t) = A \cos(\omega t + \phi) \)
In this equation:
- \( A \) is the amplitude, representing the maximum displacement from equilibrium. In our example, it is 0.15 m.
- \( \omega \) is the angular frequency, calculated as 14 rad/s.
- \( \phi \) is the phase shift, which is 0 when the object starts from rest.
Thus, the specific equation of motion in our scenario is \( y(t) = 0.15 \cos(14t) \), depicting how the mass moves over time from its initial position.
frequency of oscillation
The frequency of oscillation refers to how often the mass completes a full cycle of motion in one second. It is denoted by \( f \) and is directly related to the angular frequency \( \omega \) by the formula:
- \( f = \frac{\omega}{2\pi} \)