Problem 14 - Fundamentals of Physics

Question

The equation of a transverse wave on a string is $$ y=(2.0 \mathrm{~mm}) \sin \left[\left(20 \mathrm{~m}^{-1}\right) x-\left(600 \mathrm{~s}^{-1}\right) t\right] $$ The tension in the string is \(15 \mathrm{~N}\). (a) What is the wave speed? (b) Find the linear density of this string in grams per meter.

Step-by-Step Solution

Answer

(a) Wave speed is 30 m/s. (b) Linear density is 16.7 g/m.

Step 1: Identify the wave equation components

The wave equation is given by \( y = (2.0 \mathrm{~mm}) \sin[(20 \mathrm{~m}^{-1}) x - (600 \mathrm{~s}^{-1}) t] \). From this, identify the wave number \( k = 20 \mathrm{~m}^{-1} \) and angular frequency \( \omega = 600 \mathrm{~s}^{-1} \).

Step 2: Calculate the wave speed

The wave speed \( v \) can be calculated using the relationship \( v = \frac{\omega}{k} \). Substitute the values of \( \omega = 600 \mathrm{~s}^{-1} \) and \( k = 20 \mathrm{~m}^{-1} \) to find \( v = \frac{600}{20} = 30 \mathrm{~m/s} \). Thus, the wave speed is \( 30 \mathrm{~m/s} \).

Step 3: Use wave speed formula with tension and linear density

The wave speed is also related to the tension \( T \) and linear density \( \mu \) by the formula \( v = \sqrt{\frac{T}{\mu}} \). We have \( v = 30 \mathrm{~m/s} \) and \( T = 15 \mathrm{~N} \).

Step 4: Solve for the linear density

Rearrange the formula to solve for \( \mu \): \( \mu = \frac{T}{v^2} \). Substitute \( T = 15 \mathrm{~N} \) and \( v = 30 \mathrm{~m/s} \) to find \( \mu = \frac{15}{30^2} = 0.0167 \mathrm{~kg/m} \).

Step 5: Convert linear density to grams per meter

Convert the linear density from \( \mathrm{kg/m} \) to \( \mathrm{g/m} \) by multiplying by 1000. So, \( \mu = 0.0167 \mathrm{~kg/m} \times 1000 = 16.7 \mathrm{~g/m} \).

Key Concepts

Transverse Wave

A transverse wave is a wave where the particles of the medium move perpendicular to the direction of the wave's propagation. Imagine a string being shaken up and down; the wave travels horizontally along the string, but the particles of the string move vertically. This up-and-down motion is characteristic of transverse waves.

In the context of the wave equation provided, the transverse wave is represented mathematically. The sine function in the equation indicates the periodic and oscillatory nature of these waves. This equation potentially simplifies understanding how waves behave and interact with their environment. Understanding transverse waves is crucial in physics as they model many physical phenomena, such as light and water waves.

Wave Equation

The wave equation for this particular problem is given by: \[ y=(2.0 \mathrm{~mm}) \sin \left[\left(20 \mathrm{~m}^{-1}\right) x-\left(600 \mathrm{~s}^{-1}\right) t\right] \] This equation describes how the wave's displacement varies with both position \(x\) and time \(t\). Let's break down its components:

• Amplitude (2.0 mm): The maximum displacement of the wave from its rest position.
• Wave number \( k = 20 \mathrm{~m}^{-1} \): This represents the number of wave cycles per meter and is related to the wavelength by \( k = \frac{2\pi}{\lambda} \).
• Angular frequency \( \omega = 600 \mathrm{~s}^{-1} \): This measures how many oscillations occur in one second and is connected to the wave's frequency as \( \omega = 2\pi f \).

This detailed breakdown helps in calculating various wave parameters like speed. Recognizing these components and how they connect is crucial for solving wave-related problems.

Linear Density

Linear density is defined as the mass per unit length of an object. For a string, it is denoted by \( \mu \). It plays a vital role in determining the wave speed along the string. The formula connecting wave speed \( v \), tension \( T \), and linear density \( \mu \) is:\[ v = \sqrt{\frac{T}{\mu}} \] In the exercise, the wave speed \( v \) of 30 m/s and tension \( T \) of 15 N are given. To solve for linear density, rearrange the equation:\[ \mu = \frac{T}{v^2} \] By substituting \( T = 15\, \mathrm{N} \) and \( v = 30\, \mathrm{m/s} \), we calculate \( \mu = 0.0167\, \mathrm{kg/m} \). Converting this into grams per meter results in a linear density of 16.7 g/m.
Understanding linear density makes it easier to predict how different strings will behave under tension when waves travel through them.

Tension in String

Tension in a string is the force that is applied along its length. It is measured in Newtons (N) and affects how fast a wave can travel through the string. In our example, the tension of the string is given as 15 N. This force stretches the string and influences its physical properties, especially the wave's speed.

Using the formula for wave speed\[ v = \sqrt{\frac{T}{\mu}} \]we see that increased tension results in a higher wave speed if the linear density remains constant. Understanding tension helps in grasping how waves transmit energy through mediums like strings. Tension impacts not only speed but also influences music instruments, engineering designs, and many practical applications.