Problem 13 - College Physics

Question

\(\bullet\) Transverse waves are traveling on a long string that is under a tension of 4.00 \(\mathrm{N}\) . The equation describing these waves is $$y(x, t)=(1.25 \mathrm{cm}) \sin \left[\left(415 \mathrm{s}^{-1}\right) t-\left(44.9 \mathrm{m}^{-1}\right) x\right]$$ Find the linear mass density of this string.

Step-by-Step Solution

Answer

The linear mass density is approximately 0.0468 kg/m.

Step 1: Identify Parameters from Equation

First, notice that the wave equation is given by \[ y(x, t) = (1.25 \,\mathrm{cm}) \sin\left[\left(415 \, \mathrm{s}^{-1}\right) t - \left(44.9\, \mathrm{m}^{-1}\right) x\right] \]This equation is in the general form \[ y(x, t) = A \sin(kx - \omega t) \]where the wave's angular frequency \(\omega\) is 415 \(\mathrm{s}^{-1}\) and the wave number \(k\) is 44.9 \(\mathrm{m}^{-1}\).

Step 2: Recall Relationship Between Wave Speed, Tension, and Linear Mass Density

The speed \(v\) of a wave on a string is related to the tension \(T\) and the linear mass density \(\mu\) by the equation:\[ v = \sqrt{\frac{T}{\mu}} \].

Step 3: Find the Wave Speed Using Angular Frequency and Wave Number

The speed \(v\) of the wave can also be expressed in terms of angular frequency \(\omega\) and wave number \(k\) as:\[ v = \frac{\omega}{k} \]Substitute the known values:\[ v = \frac{415 \, \mathrm{s}^{-1}}{44.9 \, \mathrm{m}^{-1}} \approx 9.242 \frac{\mathrm{m}}{\mathrm{s}} \].

Step 4: Solve for Linear Mass Density

Rearrange the relationship for wave speed\[ v = \sqrt{\frac{T}{\mu}} \]to solve for the linear mass density \(\mu\):\[ \mu = \frac{T}{v^2} \].Plug in the values:\[ \mu = \frac{4.00 \, \mathrm{N}}{(9.242 \, \mathrm{m/s})^2} \approx 0.0468 \, \mathrm{kg/m} \].

Key Concepts

Wave Speed

Wave speed is a fundamental concept in the study of waves traveling through different mediums. It represents the distance traveled by a wave per unit of time. Understanding wave speed is pivotal when analyzing how quickly waves can propagate through a given material.

In the context of a string under tension, wave speed \(v\) is determined by the properties of the string: the tension \(T\) and the linear mass density \(\mu\). Specifically, the wave speed is calculated using the formula \(v = \sqrt{\frac{T}{\mu}}\).

This relationship highlights that increasing the tension of the string increases the wave speed, while increasing the linear mass density decreases the wave speed. Remember, wave speed is crucial for determining other wave characteristics like frequency and wavelength.

Angular Frequency

Angular frequency, represented by the symbol \(\omega\), is a measure of how fast a wave oscillates in terms of radians per second. It is an essential parameter for characterizing waves, especially periodic ones, and can also be thought of as the wave’s rotational speed.

To calculate angular frequency, you utilize the relationship \(\omega = 2\pi f\), where \(f\) is the frequency of the wave in hertz. In our given wave equation, \(\omega\) is explicitly provided as 415 \(\mathrm{s}^{-1}\).

Understanding angular frequency helps you in analyzing how many cycles of the wave occur in a specified unit of time. It is closely related to other parameters like wave speed and wavelength, providing a comprehensive description of oscillatory motion in wave mechanics.

Wave Number

Wave number, denoted by \(k\), is a fundamental concept when studying wave properties. It represents the number of wave cycles in a unit length and provides information about the spatial extent of the wave. Wave number is usually given in units of \(\mathrm{m}^{-1}\).

For a wave traveling through a medium, the wave number is related to the wavelength \(\lambda\) using the equation \(k = \frac{2\pi}{\lambda}\). In the wave equation presented, the wave number \(k\) is given as 44.9 \(\mathrm{m}^{-1}\), indicating how the wave repeats along its path.

Wave number also allows us to calculate wave speed when combined with angular frequency using the equation \(v = \frac{\omega}{k}\). This relationship underscores how wave characteristics are intertwined, each affecting the other.

Wave Equation

The wave equation is a powerful mathematical representation that describes how waves propagate through different media. Typically, it can take the form \(y(x, t) = A \sin(kx - \omega t)\), where \(A\) is the amplitude, \(k\) is the wave number, and \(\omega\) is the angular frequency.

In this particular case, the wave equation \(y(x, t) = (1.25 \mathrm{cm}) \sin\left[(415 \mathrm{s}^{-1}) t - (44.9\mathrm{m}^{-1}) x\right]\) specifies a transverse wave moving along a string. This equation tells us the displacement \(y\) of the wave at any point \(x\) at any time \(t\).

The equation is essential because it allows for the prediction and visualization of wave behavior, understanding both the spatial and temporal aspects of the wave’s motion. Each component of the wave equation contributes to the complete dynamic picture of wave propagation.

Transverse Waves

Transverse waves are a type of wave where the oscillation is perpendicular to the direction of the wave’s travel. These waves are common in scenarios such as waves on a string or electromagnetic waves.

A distinctive characteristic of transverse waves is how the particles of the medium move. As the wave travels horizontally, the particles oscillate vertically. This creates crests and troughs that are peaks and valleys of wave propagation.

Transverse waves contrast with longitudinal waves, where the particle movement is parallel to wave travel. With transverse waves, understanding the direction of displacement and wave motion is crucial for applications like understanding the mechanics of light or the vibrations on a guitar string.