Problem 81 - Fundamentals of Physics

Question

A sinusoidal transverse wave traveling in the negative direction of an \(x\) axis has an amplitude of \(1.00 \mathrm{~cm}\), a frequency of \(550 \mathrm{~Hz}\), and a speed of \(330 \mathrm{~m} / \mathrm{s}\). If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (a) \(y_{m},(\) b) \(\omega,(\mathrm{c}) k\), and \((\mathrm{d})\) the correct choice of sign in front of \(\omega ?\)

Step-by-Step Solution

Answer

(a) 0.01 m, (b) 1100π rad/s, (c) 10π/3 m⁻¹, (d) use '+' in front of ω.

Step 1: Determine Amplitude, y_m

The amplitude, denoted by \( y_m \), is given directly in the problem statement. The amplitude of the wave is the maximum displacement from the equilibrium position.Given: Amplitude \( = 1.00 \, \text{cm} \).So, \( y_m = 1.00 \, \text{cm} = 0.01 \, \text{m} \).

Step 2: Calculate Angular Frequency, ω

The angular frequency \( \omega \) is related to the frequency \( f \) of the wave by the formula:\[ \omega = 2 \pi f \]Given: Frequency \( f = 550 \, \text{Hz} \).Substituting the given frequency:\[ \omega = 2 \pi \times 550 = 1100 \pi \, \text{rad/s} \]

Step 3: Calculate Wave Number, k

The wave number \( k \) is related to the wavelength \( \lambda \), which in turn is derived from the wave speed \( v \) and the frequency \( f \), using the formula:\[ v = f \lambda \]Rearranging the equation to solve for wavelength:\[ \lambda = \frac{v}{f} = \frac{330}{550} \, \text{m} \approx 0.6 \, \text{m} \]Now, \( k \) can be calculated using the formula:\[ k = \frac{2\pi}{\lambda} \]Substituting \( \lambda = 0.6 \, \text{m} \):\[ k = \frac{2\pi}{0.6} = \frac{10\pi}{3} \, \text{m}^{-1} \]

Step 4: Determine the Sign for ω Term

The problem statement indicates that the wave is traveling in the negative \( x \)-direction.The general wave equation is:\[ y(x, t) = y_m \sin(kx \mp \omega t) \]For a wave traveling in the negative \( x \)-direction, we use the positive sign with \( \omega t \):\[ y(x, t) = y_m \sin(kx + \omega t) \]

Key Concepts

Sinusoidal Wave

Sinusoidal waves are a fundamental concept in physics and engineering. These waves have a distinct pattern that resembles the sine function from trigonometry. They are periodic, meaning they repeat at regular intervals, and they possess properties such as amplitude, frequency, wavelength, and phase. For instance, in the problem provided, the sinusoidal wave travels in the negative direction of the x-axis with an amplitude of 1.00 cm. This amplitude is the maximum extent of the wave's displacement from its equilibrium or rest position. Sinusoidal waves like this one are prevalent in nature and can describe various physical phenomena such as sound waves, light waves, and even alternating current in electricity.

Angular Frequency

Angular frequency, symbolized by \( \omega \), illustrates how quickly the wave oscillates over time. Interestingly, angular frequency is closely tied to the standard frequency \( f \), which is measured in Hertz (Hz), by a simple but powerful relationship: \( \omega = 2 \pi f \). This relationship emerges from the cyclical nature of waves, translating the frequency from cycles per second to radians per second. This conversion is crucial when dealing with sinusoidal wave equations, as it allows for precise description of the oscillatory motion.
For example, in the exercise, the frequency was given as 550 Hz. Utilizing our relation, the angular frequency was computed as \( 1100 \pi \, \text{rad/s} \). This calculation aligns the temporal aspect of the wave's oscillation to its sinusoidal formula, marking each complete cycle with a rotation of \( 2\pi \) radians.

Wave Number

The wave number, denoted by \( k \), is another pivotal characteristic of waves. It provides insight into the wave's spatial properties, essentially telling us how the sinusoidal pattern repeats over distance. The wave number is calculated using the formula \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength. Wavelength is the distance over which the wave's shape repeats.
To determine the wave number, first find the wavelength by dividing the wave speed \( v \) by the frequency \( f \). In this context, the problem gave a wave speed of 330 m/s and a frequency of 550 Hz, resulting in a wavelength of approximately 0.6 m. From there, the wave number was calculated as \( \frac{10\pi}{3} \) \( \text{m}^{-1} \). The wave number thus helps relate the spatial content of the wave in direct connection with its physical makeup.

Wave Speed

Wave speed is an essential parameter that describes how fast the wave propagates through space. It is determined by the relationship between the wavelength \( \lambda \) and the frequency \( f \): \( v = f \lambda \). This equation beautifully captures how the speed is a product of how long the wave is and how often it oscillates per unit time.
In the problem, the wave speed was given as 330 m/s. With this known, the wavelength was deduced using the prior formula, granting a practical understanding of the wave's moving characteristics through the associated frequency. Wave speed is crucial in multiple applications ranging from understanding seismic waves in geology to analyzing sound waves in acoustics. Such applications show the prominent role wave dynamics play across diverse scientific disciplines.