Problem 43 - College Physics

Question

You've made a simple pendulum with a length of \(1.55 \mathrm{~m}\), and you also have a (very light) spring with force constant \(2.45 \mathrm{~N} / \mathrm{m} .\) What mass should you add to the spring so that its period will be the same as that of your pendulum?

Step-by-Step Solution

Answer

The mass should be approximately 0.387 kg.

Step 1: Understand the Problem

In this exercise, we're comparing the periods of a pendulum and a spring-mass system. We need to find the mass that should be added to a spring for its period to match that of a pendulum with a given length.

Step 2: Pendulum Period Formula

The period of a simple pendulum is given by the formula \( T_p = 2\pi \sqrt{\frac{L}{g}} \),where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity \( (9.81 \text{ m/s}^2) \).

Step 3: Calculate Pendulum Period

Using the values given, \( L = 1.55 \text{ m} \) and \( g = 9.81 \text{ m/s}^2 \), we calculate:\[ T_p = 2\pi \sqrt{\frac{1.55}{9.81}} \approx 2\pi \times 0.397 \approx 2.50\text{ s}. \]

Step 4: Spring-Mass System Period Formula

The period of a mass-spring system is given by the formula \( T_s = 2\pi \sqrt{\frac{m}{k}} \),where \( m \) is the mass, and \( k \) is the spring constant.

Step 5: Set Equations Equal and Solve for Mass

To find the mass \( m \) that makes the periods equal \( (T_p = T_s) \), we set \[ 2\pi \sqrt{\frac{1.55}{9.81}} = 2\pi \sqrt{\frac{m}{2.45}}. \]Solving for \( m \), we have:\[ \frac{1.55}{9.81} = \frac{m}{2.45}. \]Cross-multiply and solve for \( m \):\[ m = \frac{1.55 \times 2.45}{9.81} \approx 0.387 \text{ kg}. \]

Step 6: Confirm the Solution

Verify that calculations are consistent and meet the problem's requirements. Check to ensure no arithmetic or algebraic errors are present.

Key Concepts

Period of Pendulum

The period of a pendulum is the time it takes for one complete swing, back and forth. For a simple pendulum, this period depends mostly on the length of the pendulum and the acceleration due to gravity. The equation that describes this is:\[T_p = 2\pi \sqrt{\frac{L}{g}}\]where:

• \(T_p\) is the period of the pendulum.
• \(L\) is the length of the pendulum.
• \(g\) is the acceleration due to gravity, typically \(9.81 \text{ m/s}^2\) on the surface of the Earth.

In this particular problem, a simple pendulum has a length of \(1.55 \text{ m}\). Plugging this into the formula, the calculated period is approximately \(2.50\) seconds. This means that every full swing of the pendulum takes \(2.50\) seconds to complete.

Spring-Mass System

A spring-mass system consists of a mass attached to a spring. When displaced, the system oscillates in simple harmonic motion. The time it takes to complete one full cycle of oscillation is called its period, \(T_s\). The formula for the period of a spring-mass system is given by:\[T_s = 2\pi \sqrt{\frac{m}{k}}\]where:

• \(T_s\) is the period of the spring-mass system.
• \(m\) is the mass attached to the spring.
• \(k\) is the spring constant, which describes the stiffness of the spring.

To find the mass required to match the period of our pendulum, we need to solve for \(m\) when \(T_s = 2.50 \text{ s}\). The given spring constant is \(2.45 \text{ N/m}\). By equating both periods and solving for mass, we ensure that the oscillation of the spring-mass system matches the pendulum's oscillation.

Harmonic Motion

Harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement. Both the simple pendulum and spring-mass system exhibit harmonic motion. This means they both move in repeating cycles and have predictable behaviors.

Characteristics of Harmonic Motion

• **Restoring Force:** In a pendulum, gravity provides the restoring force, while in a spring-mass system, it is the spring force.
• **Predictable Cycles:** The systems return to the starting point after a period, completing a cycle.
• **Dependence on Physical Characteristics:** The period of motion depends on elements like length or mass, not on how big the swings are.

Understanding harmonic motion is crucial for calculating and predicting the cycles and behaviors of pendulums and spring-mass systems. This assists in determining identical periods by adjusting mass or length, depending on the system.

Mass Calculation

Calculating mass for a spring-mass system involves algebraically manipulating the period formula. For this exercise, we set the pendulum period equal to the spring-mass system period:\[2\pi \sqrt{\frac{1.55}{9.81}} = 2\pi \sqrt{\frac{m}{2.45}}\]After removing \(2\pi\) from both sides, our equation becomes:\[\sqrt{\frac{1.55}{9.81}} = \sqrt{\frac{m}{2.45}}\]By squaring both sides, we simplify it to:\[\frac{1.55}{9.81} = \frac{m}{2.45}\]Cross-multiplying, we find the mass \(m\):\[m = \frac{1.55 \times 2.45}{9.81} \approx 0.387 \text{ kg}\]This calculation ensures that the spring oscillates in harmony with our pendulum, having the same period. It's the perfect balance for equal rhythm in both systems.