Problem 57 - Fundamentals of Physics

Question

A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(8.00 \mathrm{~s}^{-1}\right) t\right] $$ and a generator at the other end creates the wave $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x-\left(8.00 \mathrm{~s}^{-1}\right) t\right] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For \(x \geq 0\), what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of \(x\) ? For \(x \geq 0\), what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value of \(x\) ?

Step-by-Step Solution

Answer

Frequency: 1.27 Hz; Wavelength: \(\pi\) m; Speed: 4.00 m/s. Nodes at 0, 1.57, 3.14 m; Antinodes at 0.785, 2.36, 3.93 m.

Step 1: Finding the Frequency

The frequency \(f\) of a wave can be determined from the angular frequency \(\omega\) provided in the wave equation. Here \(\omega = 8.00 \, \mathrm{s}^{-1}\). The relation between angular frequency and frequency is \(\omega = 2\pi f\). Therefore, \[ f = \frac{\omega}{2\pi} = \frac{8.00}{2\pi} \approx 1.27 \, \mathrm{Hz}. \] The frequency of each wave is approximately \(1.27 \, \mathrm{Hz}\).

Step 2: Calculating the Wavelength

The wave number \(k\) in the wave equation is \(2.00 \, \mathrm{m}^{-1}\). Wavelength \(\lambda\) is related to the wave number by \(k = \frac{2\pi}{\lambda}\). Solving for \(\lambda\), we find: \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{2.00} = \pi \, \mathrm{m}. \] The wavelength of each wave is \(\pi \, \mathrm{m}\).

Step 3: Determining the Speed of the Wave

Wave speed \(v\) can be found using the relationship \(v = f\lambda\). Substituting the values of \(f\) and \(\lambda\), \[ v = 1.27 \times \pi \approx 4.00 \, \mathrm{m/s}. \] The speed of each wave is approximately \(4.00 \, \mathrm{m/s}\).

Step 4: Finding the Location of the First Node

For the given standing wave pattern, nodes occur where the displacement is always zero. These occur at \(x_n = n \frac{\lambda}{2}, n = 0, 1, 2, \ldots \). Given \(\lambda = \pi \, \mathrm{m}\), the smallest node (\(n=0\)) is at \(x_0 = 0 \, \mathrm{m}\).

Step 5: Finding the Location of the Second Node

The second smallest node location (\(n=1\)) is at \(x_1 = 1 \frac{\lambda}{2} = \frac{\pi}{2} \approx 1.57 \, \mathrm{m}\).

Step 6: Finding the Location of the Third Node

The third smallest node location (\(n=2\)) is at \(x_2 = 2 \frac{\lambda}{2} = \pi \approx 3.14 \, \mathrm{m}\).

Step 7: Finding the Location of the First Antinode

Antinodes occur at positions of maximum displacement, located at \(x_a = (n + 0.5)\frac{\lambda}{2}\). The smallest antinode (\(n=0\)) is at \(x_a = 0.5 \frac{\pi}{2} = \frac{\pi}{4} \approx 0.785 \, \mathrm{m}\).

Step 8: Finding the Location of the Second Antinode

The second smallest antinode location (\(n=1\)) is at \(x_a = 1.5 \frac{\pi}{2} = \frac{3\pi}{4} \approx 2.36 \, \mathrm{m}\).

Step 9: Finding the Location of the Third Antinode

The third smallest antinode location (\(n=2\)) is at \(x_a = 2.5 \frac{\pi}{2} = \frac{5\pi}{4} \approx 3.93 \, \mathrm{m}\).

Key Concepts

Frequency Calculation

Understanding the frequency of a wave involves determining how many cycles or oscillations the wave completes in one second. In the problem you are working on, the angular frequency, denoted as \(\omega\), is \(8.00 \, \mathrm{s}^{-1}\). The frequency \(f\) can be calculated using the formula: \[ f = \frac{\omega}{2\pi} \].
This formula comes from the relationship between angular frequency \(\omega\) and frequency \(f\), where \(\omega = 2\pi f\) ties the two together.
Once you plug in the provided angular frequency, the calculation becomes straightforward:

• \( f = \frac{8.00}{2\pi} \approx 1.27 \, \mathrm{Hz} \)

Hence, the frequency indicates how rapidly the wave oscillates, giving us approximately 1.27 cycles per second.

Wavelength Determination

The wavelength of a wave, represented as \(\lambda\), is the distance between successive peaks of the wave. This can be calculated using the wave number \(k\), given as \(2.00 \, \mathrm{m}^{-1}\). The relationship is described as \(k = \frac{2\pi}{\lambda}\). Solving for \(\lambda\), you obtain:
\[ \lambda = \frac{2\pi}{k} \]
By substituting \(k = 2.00 \, \mathrm{m}^{-1}\), it yields:

• \( \lambda = \frac{2\pi}{2.00} = \pi \, \mathrm{m} \)

This means the distance over which the wave's shape repeats itself is \(\pi \, \mathrm{meters}\). The straightforward nature of this calculation makes identifying wave characteristics easier in physical systems.

Wave Speed

Wave speed is a measure of how fast the wave propagates through a medium. It's determined through the equation \(v = f\lambda\), where \(v\) is the wave speed, \(f\) is the frequency, and \(\lambda\) is the wavelength.
From previous calculations, we know the frequency \(f \approx 1.27 \, \mathrm{Hz}\) and the wavelength \(\lambda = \pi \, \mathrm{m}\). Therefore, the speed \(v\) can be calculated as:

• \( v = 1.27 \times \pi \approx 4.00 \, \mathrm{m/s} \)

The wave moves at approximately 4 meters per second, demonstrating how frequency and wavelength together describe wave motion efficiently.

Standing Wave Nodes

Nodes in a standing wave are points along the medium where there is no movement—the wave amplitude at these positions is always zero. This happens at regular intervals along the string, calculated using the formula \(x_n = n \frac{\lambda}{2}\), where \(n\) is a whole number (0, 1, 2, …).
For the smallest nodes with the given wavelength \(\lambda = \pi \, \mathrm{m}\):

• First node: \(x_0 = 0 \, \mathrm{m}\)
• Second node: \(x_1 = \frac{\pi}{2} \approx 1.57 \, \mathrm{m}\)
• Third node: \(x_2 = \pi \approx 3.14 \, \mathrm{m}\)

Knowing the nodes helps in understanding points of destructive interference, where wave crests and troughs cancel each other out.

Antinodes Location

Antinodes are points of maximum displacement in a standing wave, sitting midway between nodes. At these locations, the wave's amplitude is greatest. They are calculated using \(x_a = \left(n + 0.5\right)\frac{\lambda}{2}\).
Using the same wavelength \(\lambda = \pi \, \mathrm{m}\), the antinodes are positioned at:

• First antinode: \(x_{a,0} = \frac{\pi}{4} \approx 0.785 \, \mathrm{m}\)
• Second antinode: \(x_{a,1} = \frac{3\pi}{4} \approx 2.36 \, \mathrm{m}\)
• Third antinode: \(x_{a,2} = \frac{5\pi}{4} \approx 3.93 \, \mathrm{m}\)

Antinode placement is crucial for understanding constructive interference within the wave, where peaks align to enhance oscillations.