Problem 82 - College Physics

Question

Spring \(\mathrm{A}(50.0 \mathrm{~N} / \mathrm{m})\) is attached to the ceiling. The top of spring \(\mathrm{B}(30.0 \mathrm{~N} / \mathrm{m})\) is hooked onto the bottom of spring A. Then a 0.250-kg mass is then attached to the bottom of Spring B. (a) How far will the object fall until it reaches equilibrium? (b) What is the period of the resulting oscillation?

Step-by-Step Solution

Answer

(a) 0.1307 m, (b) 0.726 s

Step 1: Identify forces at equilibrium

At equilibrium, the force of gravity on the mass equals the combined restoring force of the two springs. This means the gravitational force \( F_g = m \cdot g \) should equal the total spring force, which can be modeled by adding the spring constants because they are in series: \( k_{total} = \frac{1}{\frac{1}{k_A} + \frac{1}{k_B}} \). Calculate \( F_g \) using \( m = 0.250 \text{ kg} \) and \( g = 9.8 \, \text{m/s}^2 \).

Step 2: Calculate total spring constant

For springs in series, the equivalent spring constant \( k_{total} \) is given by the reciprocal of the sum of the reciprocals of the individual spring constants: \[ \frac{1}{k_{total}} = \frac{1}{k_A} + \frac{1}{k_B} \]. Here, \( k_A = 50.0 \, N/m \) and \( k_B = 30.0 \, N/m \). Plug these values into the formula to find \( k_{total} \).

Step 3: Determine displacement at equilibrium

Now, use Hooke's Law which states \( F = k \cdot x \) where \( F \) is the force and \( x \) is the displacement. Set \( F_g = k_{total} \cdot x \) and solve for \( x \) using the previously calculated \( F_g \) and \( k_{total} \).

Step 4: Find the period of oscillation

The period of oscillation for a mass attached to a spring system is given by \( T = 2\pi \sqrt{\frac{m}{k_{total}}} \). Use the mass \( m = 0.250 \text{ kg} \) and the total spring constant \( k_{total} \) calculated earlier to find \( T \).

Step 5: Solve equations

Using the formulas and values from previous steps:1. For the total spring constant:\[ \frac{1}{k_{total}} = \frac{1}{50.0} + \frac{1}{30.0} = \frac{3}{150} + \frac{5}{150} = \frac{8}{150} \implies k_{total} = \frac{150}{8} = 18.75 \, N/m \]2. For the displacement at equilibrium:\[ F_g = 0.250 \times 9.8 = 2.45 \, N \]\[ 2.45 = 18.75 \times x \implies x = \frac{2.45}{18.75} = 0.1307 \, m \]3. For the period of the resulting oscillation:\[ T = 2\pi \sqrt{\frac{0.250}{18.75}} = 2\pi \times 0.1155 = 0.726 \, s \]

Step 6: Final Results

(a) The object will fall 0.1307 meters until it reaches equilibrium. (b) The period of the resulting oscillation is 0.726 seconds.

Key Concepts

Equilibrium in Physics

Equilibrium in physics describes a condition where all the forces acting on a system are balanced. In this exercise, equilibrium is achieved when the gravitational force on the mass is exactly balanced by the restoring force of the two springs combined. When in equilibrium, the system is at rest and not accelerating.

For a mass suspended by springs in the problem, we start by calculating the gravitational force using the formula:

• The force of gravity is calculated by: \( F_g = m \cdot g \), where \( m \) is the mass (0.250 kg) and \( g \) is the acceleration due to gravity (9.8 m/s²).

It is important to understand that every point in the problem behaves according to this balancing act. When forces are at equilibrium, even if the entire system is not moving, subtle internal tensions between the forces tell us a lot about the system's potential energy and stability.

Hooke's Law

Hooke's Law is a fundamental principle that explains the behavior of springs. It tells us that the force exerted by a spring is proportional to the distance it is compressed or stretched from its rest position. This relationship can be expressed as:

• \( F = k \cdot x \), where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium.

In the context of the problem, when the system reaches equilibrium, the gravitational force is equal to the total spring force. Using Hooke's Law helps us find out how much the spring extends when the weight is added. Keep in mind that this law applies perfectly only within the elastic limit of the spring, meaning the point before any permanent deformation occurs.

This relationship is essential for predicting how materials will respond to forces in various engineering applications, as it can inform the limits within which the material can safely operate.

Spring Constant

The spring constant, represented by \( k \), is a measure of a spring's stiffness. Higher values of \( k \) indicate a stiffer spring that requires more force to produce a given amount of stretch or compression.

When dealing with series spring systems like in this problem, we must calculate an equivalent spring constant, \( k_{total} \), because the two springs work together to support the load. The formula for springs in series is:

• \( \frac{1}{k_{total}} = \frac{1}{k_A} + \frac{1}{k_B} \)

This formula comes from the fact that each spring extends a little bit to share the load, and the total extension is influenced by both springs. It's crucial to use the proper method of finding a combined spring constant for series arrangements to ensure the accuracy of your calculations.

In practical situations, understanding the spring constant allows us to design systems that can safely handle predetermined loads and maintain desired performance levels.

Period of Oscillation

The period of oscillation refers to the time it takes for a complete cycle of motion. For a mass-spring system, this is influenced by both the mass and the spring constant of the system. The formula for calculating this period, \( T \), is:

• \( T = 2\pi \sqrt{\frac{m}{k_{total}}} \), where \( m \) is the mass and \( k_{total} \) is the spring constant calculated for the system.

This formula tells us that heavier objects or softer springs result in longer periods, meaning it takes more time for the system to complete one cycle of motion.

Knowing the period of oscillation can be useful in designing mechanical systems, such as car suspensions or seismic isolators, where controlling the timing and magnitude of movements can impact overall performance and safety. In this exercise, calculating the period helps illustrate how the mass and the spring work together to produce oscillating motion, providing a dynamic aspect to equilibrium analysis.