Problem 84 - University Physics with Modern Physics

Question

cp Longitudinal Waves on a Spring. A long spring such as a Slinky"is is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke's law has mass \(m,\) length \(L,\) and force constant \(k^{\prime},\) the speed of longitudinal waves on the spring is \(v=L \sqrt{k^{\prime} / m}\) . (see Section 16.2\()\) (b) Evaluate \(v\) for a spring with \(m=0.250 \mathrm{kg}, L=2.00 \mathrm{m},\) and \(k^{\prime}=1.50 \mathrm{N} / \mathrm{m}\)

Step-by-Step Solution

Answer

The speed of the longitudinal wave is approximately 4.90 m/s.

Step 1: Understand Hooke's Law

Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, this can be expressed as \( F = -k\Delta x \), where \(k\) is the spring constant and \(\Delta x\) is the displacement from the equilibrium position.

Step 2: Relate Wave Speed to Spring Properties

The speed of a wave on a spring, when the spring obeys Hooke's Law, is given by the mechanical wave speed formula \( v = \sqrt{\frac{k^{\prime}L}{m}} \). This formula is derived using the properties of the spring: tension provided by the spring constant \(k^{\prime}\), the mass \(m\), and length \(L\).

Step 3: Evaluate the Expression for v

We need to show \( v = L \sqrt{\frac{k^{\prime}}{m}} \). Start with \( v = \sqrt{\frac{k^{\prime} L}{m}} \) and recognize that this expression can be rewritten as \( v = L \cdot \sqrt{\frac{k^{\prime}}{m}} \). This shows how length and the ratio of spring constant to mass determine speed.

Step 4: Plug in Given Values to Calculate v

Given \( m = 0.250 \, \text{kg} \), \( L = 2.00 \, \text{m} \), and \( k^{\prime} = 1.50 \, \text{N/m} \), substitute these into the expression: \[ v = 2.00 \cdot \sqrt{\frac{1.50}{0.250}} = 2.00 \cdot \sqrt{6.0} \]. Calculate \( \sqrt{6.0} \approx 2.45 \), so \( v \approx 2.00 \times 2.45 = 4.90 \, \text{m/s} \).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes how springs behave when forces are applied to them in mechanics. It tells us that the force required to stretch or compress a spring by a certain length is directly proportional to that length. This relationship is mathematically represented by the formula:

• \( F = -k\Delta x \),

where:

• \( F \) is the force applied to the spring,
• \( k \) is the spring constant, representing the stiffness of the spring,
• \( \Delta x \) is the change in length from its original position.

The minus sign indicates that the force exerted by the spring is in the opposite direction of the displacement. Understanding this law is crucial, as it lays the foundation for analyzing wave speed on a spring, such as when observing longitudinal waves.

Wave Speed Formula

The speed of a wave on a spring can be calculated using the wave speed formula. This formula helps determine how quickly a longitudinal wave travels through the spring and is given by: \[ v = \sqrt{\frac{k^{\prime}L}{m}} \] In this equation:

• \( v \) is the wave speed,
• \( k^{\prime} \) is the force constant of the spring, a variant of the spring constant used in real setups,
• \( L \) is the length of the spring,
• \( m \) represents the mass of the spring.

This formula is crucial because it demonstrates how both the physical properties of the spring and Hooke's Law impact wave behavior.

Spring Constant

The spring constant \( k \) is a measure of how stiff a spring is. A higher spring constant indicates a stiffer spring, while a lower spring constant means the spring is more flexible. In the context of wave speed, the force constant \( k^{\prime} \) plays a similar role. When the spring obeys Hooke's Law, \( k^{\prime} \) contributes to the tension within the spring that influences wave speed.
This role is evident in the wave speed formula, where a higher \( k^{\prime} \) results in greater wave speeds if other factors remain constant.
Finding the right spring constant is important in designing springs for specific mechanical applications, ensuring they respond to forces as needed.

Mass

Mass \( m \) is another critical factor affecting the speed of longitudinal waves on a spring. In physics, mass describes how much matter an object contains and is typically measured in kilograms (kg). For wave propagation, the mass of the spring contributes to the inertia, or resistance to motion, of the wave. As seen in the wave speed formula \( v = \sqrt{\frac{k^{\prime}L}{m}} \), a smaller mass allows waves to move faster through the spring, while a larger mass slows them down.
This relationship highlights how mass, along with the spring constant and length, collectively impacts how energy is transmitted as a wave.

Mechanics

Mechanics is the branch of physics that studies the forces and motion of objects. It provides the framework to understand phenomena like wave propagation on a spring. Using Newton's laws of motion, mechanics examines how forces, mass, and distance interact to affect the speed and behavior of longitudinal waves.
By assimilating concepts such as Hooke's Law, the wave speed formula, and the properties of mass and spring constant, mechanics allows us to predict how a spring will react under various conditions. In practice, applying these principles helps engineers design and optimize systems that use springs, ensuring efficiency and efficacy in mechanical operations.