Problem 89 - Calculus Volume 3

Question

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?

Step-by-Step Solution

Answer

The equation of motion is \( y(t) = 0.15 \cos(14t) \), and the frequency is approximately 2.23 Hz.

Step 1: 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.

Step 2: 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:\[ k = \frac{mg}{x} = \frac{0.4 \times 9.8}{0.05} = 78.4 \, \text{N/m} \]

Step 3: Determine Angular Frequency (ω)

The angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{k}{m}} \). Using \( k = 78.4 \, \text{N/m} \) and \( m = 0.4 \, \text{kg} \):\[ \omega = \sqrt{\frac{78.4}{0.4}} = \sqrt{196} = 14 \, \text{rad/s} \]

Step 4: 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, \( \phi = 0 \):\[ y(t) = 0.15 \cos(14t) \]

Step 5: Calculate the Frequency

The frequency \( f \) is derived from angular frequency \( \omega = 2\pi f \):\[ f = \frac{\omega}{2\pi} = \frac{14}{2\pi} \approx 2.23 \, \text{Hz} \]

Key Concepts

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:

• \( F = kx \)

Here, \( F \) is the force applied to the spring, \( k \) is the spring constant, which measures the stiffness of the spring, and \( x \) is the displacement of the spring from its original position.
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:

• \( \omega = \sqrt{\frac{k}{m}} \)

In this context, \( k \) is the spring constant (which we found to be 78.4 N/m), and \( m \) is the mass, which is 0.4 kg. By applying these values, the angular frequency \( \omega \) is determined to be 14 rad/s. This tells us how rapidly the mass will oscillate back and forth along its path.

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:

• \( y(t) = A \cos(\omega t + \phi) \)

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:

• \( f = \frac{\omega}{2\pi} \)

In our example, with an angular frequency \( \omega \) of 14 rad/s, the frequency \( f \) is calculated to be approximately 2.23 Hz. This means the mass completes about 2.23 oscillations each second. Understanding the frequency helps in predicting how the system behaves over time and is crucial for designing systems involving periodic motion.