Problem 17 - Fundamentals of Physics

Question

The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m} .\) A transverse wave on the string is described by the equation $$ y=(0.021 \mathrm{~m}) \sin \left[\left(2.0 \mathrm{~m}^{-1}\right) x+\left(30 \mathrm{~s}^{-1}\right) t\right] $$ What are (a) the wave speed and (b) the tension in the string?

Step-by-Step Solution

Answer

(a) 15 m/s; (b) 0.036 N.

Step 1: Identify Given Values and Equation

The problem provides us with the linear density of the string, \( \mu = 1.6 \times 10^{-4} \ \mathrm{kg/m} \), and a wave equation which is \( y = (0.021 \ \mathrm{m}) \sin[(2.0 \ \mathrm{m}^{-1}) x + (30 \ \mathrm{s}^{-1}) t] \). From this equation, we can identify the wave number \( k = 2.0 \ \mathrm{m}^{-1} \) and the angular frequency \( \omega = 30 \ \mathrm{s}^{-1} \).

Step 2: Calculate the Wave Speed

The wave speed \( v \) in terms of angular frequency \( \omega \) and wave number \( k \) is given by the formula \( v = \frac{\omega}{k} \). Substitute the known values: \( \omega = 30 \ \mathrm{s}^{-1} \) and \( k = 2.0 \ \mathrm{m}^{-1} \) into this equation: \[ v = \frac{30}{2.0} = 15 \ \mathrm{m/s} \]. Therefore, the wave speed is \( 15 \ \mathrm{m/s} \).

Step 3: Find the Tension in the String

Using the relationship between wave speed, tension \( T \), and linear density \( \mu \), we have \( v = \sqrt{\frac{T}{\mu}} \). Square both sides: \( v^2 = \frac{T}{\mu} \). Solve for the tension \( T \): \( T = v^2 \mu \). Substitute the known values: \( v = 15 \ \mathrm{m/s} \) and \( \mu = 1.6 \times 10^{-4} \ \mathrm{kg/m} \) into the equation: \[ T = (15)^2 \times 1.6 \times 10^{-4} = 0.036 \ \mathrm{N} \]. Therefore, the tension in the string is \( 0.036 \ \mathrm{N} \).

Key Concepts

Wave Speed Calculation

Understanding wave speed is essential in wave mechanics. It describes how fast the wave travels through a medium. To calculate the wave speed for a transverse wave on a string, we use the equation: \[ v = \frac{\omega}{k} \] This links the wave's angular frequency \( \omega \) and its wave number \( k \).

• Angular frequency \( \omega \) measures how quickly the wave oscillates in radians per second.
• Wave number \( k \) represents the number of wave cycles per meter.

Given that \( \omega = 30 \, \mathrm{s}^{-1} \) and \( k = 2.0 \, \mathrm{m}^{-1} \), we substitute these values into the formula: \[ v = \frac{30}{2.0} = 15 \, \mathrm{m/s} \] Thus, the wave speed, \( v \), is \( 15 \, \mathrm{m/s} \), indicating the wave advances 15 meters each second through the string.
Recognizing these relationships helps clarify the wave's dynamics and behavior in various contexts.

Tension in a String

The tension in a string plays a critical role in determining the wave speed, impacting how the wave moves. This tension relates to how tight the string is pulled, and it influences the wave's characteristics significantly. To find the tension, \( T \), we use the formula that relates wave speed \( v \), tension \( T \), and linear density \( \mu \): \[ v = \sqrt{\frac{T}{\mu}} \] Square both sides to solve for tension: \[ v^2 = \frac{T}{\mu} \] Then, solve for \( T \): \[ T = v^2 \mu \] With known values, \( v = 15 \, \mathrm{m/s} \) and \( \mu = 1.6 \times 10^{-4} \, \mathrm{kg/m} \), the equation becomes:
\[ T = (15)^2 \times 1.6 \times 10^{-4} = 0.036 \, \mathrm{N} \] This result, \( T = 0.036 \, \mathrm{N} \), indicates the tension level within the string. Knowing this value is crucial, as it affects not only the wave speed but also the string's stability and quality of vibrations.

Linear Density of a String

Linear density, symbolized by \( \mu \), is crucial when discussing waves on strings. It describes how much mass lies along a unit length of the string and is expressed in kilograms per meter \( \mathrm{kg/m} \). A higher linear density means more mass per length, which affects how a wave travels along the string. The given linear density for the string is \( 1.6 \times 10^{-4} \, \mathrm{kg/m} \). When calculating wave characteristics, this value is key, especially when evaluating wave speed or determining tension.

• Influence on Wave Speed: In the relationship \( v = \sqrt{\frac{T}{\mu}} \), we see that wave speed \( v \) inversely relates to \( \mu \). A string with greater density travels waves more slowly.
• Influence on Tension: For the same tension, a denser string supports a slower wave speed, or needs greater tension for the same wave speed.

A thorough understanding of linear density allows us to manipulate wave properties effectively and predicts how waves respond in different materials. This concept is pivotal for perfecting tension calculations and adjusting stringed instruments or designing mechanisms involving cables or threads.