Problem 77 - Fundamentals of Physics

Question

The type of rubber band used inside some baseballs and golf balls obeys Hooke's law over a wide range of elongation of the band. A segment of this material has an unstretched length \(\ell\) and a mass \(m .\) When a force \(F\) is applied, the band stretches an additional length \(\Delta \ell .\) (a) What is the speed (in terms of \(m, \Delta \ell\), and the spring constant \(k\) ) of transverse waves on this stretched rubber band? (b) Using your answer to (a), show that the time required for a transverse pulse to travel the length of the rubber band is proportional to \(1 / \sqrt{\Delta \ell}\) if \(\Delta \ell \& \ell\) and is constant if \(\Delta \ell\) ? \(\ell\).

Step-by-Step Solution

Answer

The speed of transverse waves is \(v = \sqrt{\frac{k \Delta \ell (\ell + \Delta \ell)}{m}}\). The time is \(\propto 1/\sqrt{\Delta \ell}\) if \(\Delta \ell \ll \ell\) and constant if \(\Delta \ell \gg \ell\).

Step 1: Express the speed of a transverse wave on a stretched band

The speed \( v \) of a transverse wave on a stretched string is given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string and \( \mu \) is the linear mass density. For Hooke's law, the tension \( T \) can be expressed as \( T = k \Delta \ell \). The linear mass density \( \mu \) is given by \( \mu = \frac{m}{\ell + \Delta \ell} \).

Step 2: Substitute tension and linear mass density into the wave speed formula

Substituting the expressions for tension and linear mass density from Step 1 into the wave speed formula, we get:\[v = \sqrt{\frac{k \Delta \ell}{m/(\ell + \Delta \ell)}} = \sqrt{\frac{k \Delta \ell (\ell + \Delta \ell)}{m}}\]Simplifying this expression, we obtain:\[v = \sqrt{\frac{k \Delta \ell (\ell + \Delta \ell)}{m}}\]

Step 3: Determine the time for a transverse pulse to travel the length of the band

The time \( t \) required for a transverse pulse to travel the length of the band is \( t = \frac{L}{v} \), where \( L = \ell + \Delta \ell \) is the total stretched length of the band. Substituting the expression for \( v \) from Step 2, we get:\[t = \frac{\ell + \Delta \ell}{\sqrt{\frac{k \Delta \ell (\ell + \Delta \ell)}{m}}} = \sqrt{\frac{m}{k \Delta \ell}} \class{K \ell + \Delta \ell}\]Simplify this to find:\[t = \sqrt{\frac{m}{k \Delta \ell}} \times (\ell + \Delta \ell)\]

Step 4: Analyze the time expression for different conditions of elongation

1. If \( \Delta \ell \ll \ell \), then \( \ell + \Delta \ell \approx \ell \). The time simplifies to:\[t = \sqrt{\frac{m}{k \Delta \ell}} \cdot \ell \propto \frac{1}{\sqrt{\Delta \ell}}\]2. If \( \Delta \ell \gg \ell \), then \( \ell + \Delta \ell \approx \Delta \ell \). The time becomes:\[t = \sqrt{\frac{m}{k}}\]This shows that the time is constant and does not depend on \( \Delta \ell \).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes how elastic objects like springs and rubber bands behave when forces are applied to them. The law states that the force needed to extend or compress a spring is proportional to the distance it is stretched or compressed. In simple terms, if you pull a rubber band, the force required increases linearly with how far you stretch it. This relationship is expressed by the formula:

• \( F = k \Delta \ell \)

Here, \( F \) is the force applied to the spring or rubber band, \( k \) is the spring constant, and \( \Delta \ell \) is the change in length of the rubber band when stretched. Understanding the Proportionality
- "Proportional" means that if you double the stretching distance, the force required doubles too. - The constant \( k \) determines how stiff or stretchy the material is. A larger \( k \) means the material is harder to stretch.This law applies until the limit of proportionality, beyond which the material may stretch differently or break.

Spring Constant

The spring constant \( k \) is a measure of the stiffness of a spring or other stretchy object. It's a crucial part of Hooke’s Law and plays a significant role in determining how an object reacts to forces.Understanding \( k \)
- Units: \( k \) is measured in newtons per meter (N/m), showing how much force is needed to extend the spring by a meter.- Determination: To find \( k \), you can divide the force \( F \) applied to the spring by the extension \( \Delta \ell \): \( k = \frac{F}{\Delta \ell} \).The spring constant tells us:

• How resistant the rubber band is to being stretched. A higher \( k \) indicates a more rigid material.
• Influences wave speed: In the context of wave speed on stretched materials, a larger \( k \) results in a faster wave speed because the tension \( T = k \Delta \ell \) affects the speed.

Knowing the spring constant helps predict the behavior of materials under different forces and is essential for accurate calculations in physical applications.

Linear Mass Density

Linear mass density \( \mu \) describes mass distribution along a line, such as a string or rubber band. It is the mass per unit length of an object. For a rubber band of mass \( m \) and length \( L \), \( \mu \) is calculated as:

• \( \mu = \frac{m}{L} \)

Key Points about \( \mu \)
- Units: \( \mu \) is expressed in kilograms per meter (kg/m).- For stretched bands, \( L \) includes both the original and the extended length, making \( L = \ell + \Delta \ell \).- It is crucial in calculating wave speed: the formula for transverse wave speed \( v \) involves \( \mu \): \[ v = \sqrt{\frac{T}{\mu}} \]Linear mass density affects the wave speed negatively; a higher \( \mu \) results in a slower wave speed. The relationship shows that for the same tension, a heavier string (higher \( \mu \)) results in a slower wave. Understanding \( \mu \) is essential for grasping how different physical properties, like mass and length, influence wave behavior on elastic mediums.