Problem 75 - Fundamentals of Physics

Question

(a) What is the fastest transverse wave that can be sent along a steel wire? For safety reasons, the maximum tensile stress to which steel wires should be subjected is \(7.00 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). The density of steel is \(7800 \mathrm{~kg} / \mathrm{m}^{3} .\) (b) Does your answer depend on the diameter of the wire?

Step-by-Step Solution

Answer

The wave speed is approximately 299.57 m/s and it does not depend on the wire's diameter.

Step 1: Understand the Concept

The speed of a transverse wave in a stretched wire can be found using the formula \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the wire and \( \mu \) is the mass per unit length of the wire. The tension \( T \) is related to the stress and cross-sectional area.

Step 2: Calculate the Tension

The tension \(T\) in the wire can be found using the formula for stress: \( \text{Stress} = \frac{T}{A} \), where \(A\) is the cross-sectional area. Rearranging, \( T = \text{Stress} \times A \). The maximum tensile stress given is \( 7.00 \times 10^{8} \mathrm{~N/m^2} \). Hence, \( T = 7.00 \times 10^{8} \times A \).

Step 3: Express Mass Per Unit Length

The mass per unit length \( \mu \) is given by \( \mu = \rho \times A \), where \( \rho \) is the density \(7800 \, \text{kg/m}^3\).

Step 4: Derive the Wave Speed Formula

Substitute the given expressions for \( T \) and \( \mu \) into the wave speed formula: \[ v = \sqrt{\frac{7.00 \times 10^{8} \times A}{7800 \times A}} \]. Simplifying this expression gives \[ v = \sqrt{\frac{7.00 \times 10^{8}}{7800}} \].

Step 5: Compute the Wave Speed

Calculate the speed: \[ v = \sqrt{\frac{7.00 \times 10^{8}}{7800}} = \sqrt{89743.59} \]. Hence, \( v \approx 299.57 \, \text{m/s} \).

Step 6: Check the Dependence on Diameter

Since both the tension and mass per unit length have \( A \) (cross-sectional area) as a common factor, the diameter (or area \( A \)) cancels out. Thus, the wave speed \( v \) is independent of the diameter of the wire.

Key Concepts

Transverse Waves

Transverse waves can be thought of as waves that move perpendicular to the direction of the wave itself. Imagine a rope being flicked up and down; the waves travel along the length of the rope while moving side to side.
In the context of steel wires, transverse waves are the oscillations that occur due to vibrations within the wire. The speed at which these waves travel is crucial for understanding how quickly energy is transmitted through the wire.
This speed is determined by the material properties and the tension within the wire. Lower tension or heavier wires will lead to slower wave speeds.

Tensile Stress

Tensile stress refers to the force exerted on a material along its length, intending to stretch it. In steel wires, tensile stress plays a key role in determining the maximum limit of force application.
To calculate tensile stress, we use the formula:

• Tensile Stress \( = \frac{T}{A} \)
• Where \(T\) is the tension in the wire, and \(A\) is the cross-sectional area.

For steel wires, there is a safety threshold for tensile stress, which means it's important not to exceed a value of \(7.00 \times 10^8 \text{ N/m}^2\). Exceeding this could risk breaking or permanently deforming the wire.

Density of Steel

The density of steel is an important physical property when calculating wave speed. Density is a measure of mass per unit volume and directly influences the mass per unit length of the wire. For steel, the density is given as \(7800 \text{ kg/m}^3\).
Knowing the density, we can determine how much mass is in a given volume of steel wire, which affects how the wire responds to applied forces. Higher density means more mass within a certain volume, thus affecting the speed at which waves travel through the material.

Mass Per Unit Length

Mass per unit length, often represented as \( \mu \), is derived from the density of the material and its cross-sectional area. It provides a measure of how much mass exists in a given length of the wire.This is calculated as:

• \( \mu = \rho \times A \)
• Where \( \rho \) is the density of the material, and \( A \) is the cross-sectional area.

In problems involving wave speed, knowing the mass per unit length helps determine how quickly a wave can travel through the material. In essence, it plays a crucial role in the physics of vibrations and oscillations.