Problem 51 - University Physics with Modern Physics

Question

A transverse sine wave with an amplitude of 2.50 \(\mathrm{mm}\) and a wavelength of 1.80 \(\mathrm{m}\) travels from left to right along a long, horizontal, stretched string with a speed of 36.0 \(\mathrm{m} / \mathrm{s} .\) Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle 1.35 \(\mathrm{m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverss velocity of a particle 1.35 \(\mathrm{m}\) to the right of the origin at time \(t=0.0625 \mathrm{s}\)

Step-by-Step Solution

Answer

(a) 20 Hz, 40π rad/s, \( \frac{10\pi}{9} \) rad/m. (b) \( y(x, t) = 2.50 \times 10^{-3} \sin(\frac{10\pi}{9} x - 40\pi t + \frac{\pi}{2}) \). (f) Displacement: -0.0025 m, Velocity: 0 m/s at \( t = 0.0625 \) s.

Step 1: Calculate the Frequency

The frequency \( f \) of a wave is given by \( f = \frac{v}{\lambda} \), where \( v \) is the speed and \( \lambda \) is the wavelength.Substitute the values: \[ f = \frac{36.0 \, \text{m/s}}{1.80 \, \text{m}} = 20.0 \, \text{Hz} \]

Step 2: Determine the Angular Frequency

The angular frequency \( \omega \) is related to the frequency by \( \omega = 2\pi f \).Calculating this gives: \[ \omega = 2\pi \times 20.0 \, \text{Hz} = 40\pi \, \text{rad/s} \]

Step 3: Find the Wave Number

The wave number \( k \) is given by \( k = \frac{2\pi}{\lambda} \).Using the wavelength: \[ k = \frac{2\pi}{1.80 \, \text{m}} = \frac{10\pi}{9} \, \text{rad/m} \]

Step 4: Write the Wave Function

The wave function can be expressed as \( y(x, t) = A \sin(kx - \omega t + \phi) \).Given that the initial phase \( \phi = \frac{\pi}{2} \) because it's at maximum displacement initially, the wave function becomes:\[ y(x, t) = 2.50 \times 10^{-3} \sin \left( \frac{10\pi}{9} x - 40\pi t + \frac{\pi}{2} \right) \]

Step 5: Calculate \(y(t)\) at the Origin

At the origin \( x = 0 \), the wave function for the left end is:\[ y(0, t) = 2.50 \times 10^{-3} \sin \left( 0 - 40\pi t + \frac{\pi}{2} \right) \]Simplifying gives:\[ y(0, t) = 2.50 \times 10^{-3} \cos(40\pi t) \]

Step 6: Calculate \(y(t)\) for Point 1.35 m Right of Origin

At \( x = 1.35 \, \text{m} \), the function becomes:\[ y(1.35, t) = 2.50 \times 10^{-3} \sin \left( \frac{10\pi}{9} \times 1.35 - 40\pi t + \frac{\pi}{2} \right) \]Simplifying:\[ y(1.35, t) = 2.50 \times 10^{-3} \sin \left( 1.5\pi - 40\pi t + \frac{\pi}{2} \right) = 2.50 \times 10^{-3} \cos(40\pi t) \]

Step 7: Find Maximum Transverse Velocity

The maximum transverse velocity \( v_{\text{max}} \) is \( \omega A \).Substituting values gives:\[ v_{\text{max}} = 40\pi \cdot 2.50 \times 10^{-3} = 0.1\pi \, \text{m/s} \approx 0.314 \, \text{m/s} \]

Step 8: Transverse Displacement and Velocity at Time t=0.0625 s

For a particle at \( x = 1.35 \, \text{m} \) and \( t = 0.0625 \, \text{s} \):\[ y(1.35, 0.0625) = 2.50 \times 10^{-3} \cos(40\pi \times 0.0625) = 2.50 \times 10^{-3} \cos(2.5\pi) = 2.50 \times 10^{-3} \times (-1) = -2.50 \times 10^{-3} \, \text{m} \]To find transverse velocity \( v \):\[ v = \frac{\partial}{\partial t} y(x,t) = -\omega A \sin(kx - \omega t + \phi) \]Evaluating at the given \( t \):\[ v(1.35, 0.0625) = -40\pi \times 2.50 \times 10^{-3} \sin\left(2.5\pi \right) = 0 \, \text{m/s} \]

Step 9: Final Results Summary

The frequency is 20 Hz, angular frequency is 40π rad/s, and wave number is \( \frac{10\pi}{9} \) rad/m. The wave function is \( y(x, t) = 2.50 \times 10^{-3} \sin(\frac{10\pi}{9} x - 40\pi t + \frac{\pi}{2}) \). For a particle at 1.35 m right of origin and \( t = 0.0625 \) s, the displacement is -0.0025 m, and velocity is 0 m/s.

Key Concepts

Transverse Waves

Transverse waves are an interesting type of wave where the movement of the particles is perpendicular to the direction of the wave's travel. Imagine a string that is plucked up and down, while the wave itself moves horizontally. This is exactly how a transverse wave operates. You see this phenomenon not just in strings, but also in waves on water and even in electromagnetic waves like light. Transverse waves have key characteristics:

• Amplitude: This is the maximum height from the rest position, like how tall the wave is.
• Wavelength: The distance between two consecutive crests or troughs.
• Speed: How fast the wave moves through its medium, like along a string in this case.

In our exercise, the transverse wave travels on a horizontal string, making it a vivid example of these waves in action.

Frequency and Angular Frequency

Frequency and angular frequency are closely related but play distinct roles in describing wave motion. The frequency of a wave indicates how many cycles occur in one second, measured in Hertz (Hz). In our given problem, you find the frequency using the formula:

• \( f = \frac{v}{\lambda} \)

where \( v \) is the wave speed, and \( \lambda \) is the wavelength.

The angular frequency is the rate of rotation along a circle, effectively translating linear frequency into radian measures, given by:

• \( \omega = 2\pi f \)

This tells us how fast something goes through its oscillation in radians per second. With angular frequency, you get a deeper understanding of the wave's cyclical nature in trigonometric terms. Together, these concepts allow us to analyze wave behavior more thoroughly.

Wave Function

A wave function is a mathematical description of the wave at any given time and position. It's expressed as:

• \( y(x, t) = A \sin(kx - \omega t + \phi) \)

where:

• \( A \): Amplitude, the maximum displacement from the rest position.
• \( k \): Wave number, detailing how many wavelengths fit into a given unit of distance.
• \( \omega \): Angular frequency, representing how fast the wave oscillates in radians per second.
• \( \phi \): Phase constant, indicating the wave's initial position in its cycle.

In the context of our exercise, the wave function explains the position of each point on the string throughout the wave's travel. Knowing this function allows us to predict where points on the wave, and hence the string, will be at any given time, making it a powerful tool for analysis.

Wave Number

The wave number is another key concept in understanding wave physics. It represents the number of wavelengths that fit into a unit of distance. Mathematically, it’s defined as:

• \( k = \frac{2\pi}{\lambda} \)

where \( \lambda \) is the wavelength. The wave number provides an idea of the wave's "tightness," or conversely, how spread out it is. In our example, the calculated wave number allows us to visualize how frequently the waves repeat along the string.

The wave number is particularly useful when we need to work with sinusoidal functions in the wave function, linking the spatial aspect of a wave to its mathematical description. This helps not only in theoretical physics but also in practical applications such as acoustics and optics.