Problem 22 - College Physics

Question

\(\bullet\) How much mass should be at the end of a spring \((k=100 \mathrm{~N} / \mathrm{m})\) in order to have a period of \(2.0 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

The required mass is approximately 10.13 kg.

Step 1: Understand the Problem

We need to determine the mass that must be attached to a spring with a spring constant \( k = 100 \, \text{N/m} \) to achieve a period \( T = 2.0 \, \text{s} \). The formula that relates mass, period, and spring constant in a spring-mass system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \).

Step 2: Rearrange the Formula

We start with the equation for the period of a spring-mass system: \( T = 2\pi \sqrt{\frac{m}{k}} \). We need to solve for the mass \( m \). First, isolate \( \sqrt{\frac{m}{k}} \) by dividing both sides of the equation by \( 2\pi \): \( \frac{T}{2\pi} = \sqrt{\frac{m}{k}} \).

Step 3: Square Both Sides

To eliminate the square root, square both sides of the equation: \( \left(\frac{T}{2\pi}\right)^2 = \frac{m}{k} \). This gives us a non-square-rooted equation that we can solve for \( m \).

Step 4: Solve for Mass \( m \)

Rearrange the equation to solve for \( m \): \( m = k\left(\frac{T}{2\pi}\right)^2 \). Substitute the given values \( T = 2.0 \, \text{s} \) and \( k = 100 \, \text{N/m} \) into the equation: \( m = 100\left(\frac{2.0}{2\pi}\right)^2 \).

Step 5: Calculate the Mass

Perform the calculations: \( m = 100\left(\frac{2.0}{2\pi}\right)^2 = 100\left(\frac{1}{\pi}\right)^2 \). Use \( \pi \approx 3.1416 \) to calculate: \( m = 100\left(\frac{1}{3.1416}\right)^2 \approx 10.13 \).

Key Concepts

Spring Constant

The spring constant, often represented by the letter \( k \), is a measure of a spring's stiffness. If a spring has a higher spring constant, it means the spring is stiffer and requires more force to stretch or compress by a certain amount. The units of spring constant, \( k \), are Newtons per meter (N/m), which indicates how many Newtons of force are needed for each meter of displacement. Here's a simple way to think about it:

• Imagine a slinky versus a rubber band. The slinky is less stiff than the rubber band, meaning the rubber band has a higher spring constant.
• The spring constant comes into play when calculating dynamics of spring-mass systems, such as determining how much force is needed to stretch or compress the spring.

The spring constant is crucial in the formula for the period of oscillation for a spring-mass system, which we'll explore more below. Understanding the spring constant helps in designing systems that can handle specific loads or forces without too much deformation.

Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement. In spring-mass systems, SHM is characterized by oscillations around an equilibrium point. The restoring force here is provided by the spring, pulling the mass back to its original position when it is displaced. Key features of SHM:

• The motion is sinusoidal, meaning it follows a smooth wave-like pattern.
• The velocity and acceleration are always changing as the object moves through its cycle.
• The amplitude is the maximum extent of the motion from the equilibrium position.

In the context of the exercise, the mass attached to the spring moves in a pattern that results in regular oscillations, defined by the properties of SHM. The spring constant and the mass dictate the specifics of this motion, including the frequency and period of oscillation.

Oscillation Period

The oscillation period is the time it takes for an object in simple harmonic motion to complete one full cycle of movement back and forth. It is an essential parameter to understanding the dynamics of a spring-mass system. For a simple spring-mass system, the period \( T \) is given by the formula:\[ T = 2\pi \sqrt{\frac{m}{k}} \]This formula reveals:

• \( T \) is directly proportional to the mass \( m \); more mass increases the period, slowing the oscillation.
• \( T \) is inversely proportional to the spring constant \( k \); a stiffer spring (higher \( k \)) leads to a shorter period.

In simple terms, a heavier mass or a weaker spring results in slower, longer oscillations, while a lighter mass or a stiffer spring causes quicker, shorter oscillations. Calculating the period helps predict how long a spring-mass system takes to complete a cycle, crucial for applications ranging from mechanical watches to vehicle suspension systems.