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Waves on a Stick. A flexible stick 2.0 \(\mathrm{m}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?

Short Answer

Expert verified
The wavelengths are 4.0 m, 2.0 m, and 1.33 m for the first, second, and third harmonics, respectively.

Step by step solution

01

Understanding Nodes and Antinodes

When a stick is free to vibrate, both ends should be antinodes because they are free to move. An antinode is a point of maximum amplitude, while a node is a point of zero amplitude. In this situation, the modes of vibration (harmonics) will have antinodes at both ends.
02

First Harmonic (Fundamental Frequency)

For the first harmonic, the stick has antinodes at both ends with one node in the middle. This setup resembles half of a full wavelength fitting into the length of the stick. Therefore, the wavelength \( \lambda_1 \) for the first harmonic is given by \( \lambda_1 = 2 \times L = 4.0\, \mathrm{m} \).
03

Second Harmonic

In the second harmonic, the stick will have three antinodes and two nodes. The length of the stick will now be equal to a full wavelength. Therefore, \( \lambda_2 = L = 2.0\, \mathrm{m} \).
04

Third Harmonic

In the third harmonic, the stick will have four antinodes and three nodes, representing one and a half wavelengths fitting into the stick's length, \( L \). Thus, \( \lambda_3 = \frac{2}{3} \times L = \frac{4.0}{3}\, \mathrm{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nodes and Antinodes
When you look at a stick free to vibrate, understanding the concepts of nodes and antinodes is helpful. These are key features in physical waves, such as sound or waves on a stick. A node is a point where the wave does not move, showing zero amplitude. Meanwhile, an antinode is a point where the wave moves the most, displaying maximum amplitude.

For a stick that is not fixed at the ends and can vibrate freely, both ends become antinodes. Think of antinodes as the parts of the wave that "dance" the most. When the stick vibrates, it forms specific patterns where nodes and antinodes alternate along its length.
  • Nodes: Points of no motion (zero amplitude).
  • Antinodes: Points of maximum motion (maximum amplitude).
This pattern is crucial for understanding how harmonics, or vibrational frequencies, form on the stick. Each harmonic changes how nodes and antinodes arrange themselves.
Vibrational Modes
Vibrational modes determine how an object like a stick vibrates at different frequencies. Every mode corresponds to a harmonic, a multiple of the stick's fundamental frequency. The first harmonic is the simplest vibrational mode, often called the fundamental mode.

In this mode, there is one node in the center and antinodes at both ends. This means the stick is divided into two sections, oscillating in opposite directions.

As you move to higher harmonics, the stick can oscillate in more complex patterns. For example, the second harmonic features two nodes between the ends, resulting in three sections. Increasing the mode number corresponds to an increasing number of nodes, dividing the stick into more oscillating parts.

Understanding vibrational modes is beneficial because:
  • It explains how waves form and behave on the stick.
  • It helps calculate the wavelength of each harmonic.
Each mode creates a unique vibrational pattern, showing how waves behave differently depending on frequency.
Wavelength Calculation
Calculating wavelength for each harmonic helps us understand the vibrations on the stick better. The wavelength is the distance between repeating points on a wave, measured from any point to the next identical point (like node to node or antinode to antinode).

For the first harmonic, since the stick forms half a wavelength, you need to double the stick's length to find the full wavelength. If the stick is 2.0 meters long, the wavelength becomes 4.0 meters.

In the second harmonic, the entire stick equals one full wavelength, so its length gives you the wavelength directly—2.0 meters.

Finally, the third harmonic features one and a half wavelengths spanning the stick's length. To find this wavelength, multiply the stick's length by two-thirds, resulting in approximately 1.33 meters.
  • First Harmonic: \( \lambda_1 = 4.0\,\mathrm{m} \)
  • Second Harmonic: \( \lambda_2 = 2.0\,\mathrm{m} \)
  • Third Harmonic: \( \lambda_3 = 1.33\,\mathrm{m} \)
Understanding these calculations helps reveal the relationship between the wave's speed, its frequency, and the physical dimensions it affects, such as the length of the stick.

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Most popular questions from this chapter

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates the rope transversely at 120 \(\mathrm{Hz}\) . The other end passes over a pulley and supports a \(1.50-\mathrm{kg}\) mass. The linear mass density of the rope is 0.0550 \(\mathrm{kg} / \mathrm{m}\) . (a) What is the speed of a transverse wave on the rope? (b) What is the wavelength? (c) How would your answers to parts (a) and (b) change if the mass were increased to 3.00 \(\mathrm{kg}\) ?

With what tension must a rope with length 2.50 \(\mathrm{m}\) and mass 0.120 \(\mathrm{kg}\) be stretched for transverse waves of frequency 40.0 \(\mathrm{Hz}\) to have a wavelength of 0.750 \(\mathrm{m} ?\)

A horizontal wire is stretched with a tension of 94.0 \(\mathrm{N}\) and the speed of transverse waves for the wire is 492 \(\mathrm{m} / \mathrm{s}\) . What must the amplitude of a traveling wave of frequency 69.0 \(\mathrm{Hz}\) be in order for the average power carried by the wave to be 0.365 \(\mathrm{w}\) ?

The speed of sound in air at \(20^{\circ} \mathrm{C}\) is 344 \(\mathrm{m} / \mathrm{s} .\) (a) What is the wavelength of a sound wave with a frequency of 784 \(\mathrm{Hz}\) , corresponding to the note \(\mathrm{G}_{5}\) on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher than the note in part (a)?

Waves of Arbitrary Shape. (a) Explain why any wave described by a function of the form \(y(x, t)=f(x-v t)\) moves in the \(+x\) -direction with speed \(v .\) (b) Show that \(y(x, t)=f(x-v t)\) satisfies the wave equation, no matter what the functional form of \(f .\) To do this, write \(y(x, t)=f(u),\) where \(u=x-\) vt. Then, to take partial derivatives of \(y(x, t),\) use the chain rule: $$\begin{aligned} \frac{\partial y(x, t)}{\partial t} &=\frac{d f(u)}{d u} \frac{\partial u}{\partial t}=\frac{d f(u)}{d u}(-v) \\ \frac{\partial y(x, t)}{\partial t} &=\frac{d f(u)}{d u} \frac{\partial u}{\partial x}=\frac{d f(u)}{d u} \end{aligned}$$ (c) A wave pulse is described by the function \(y(x, t)=\) \(D e^{-(B x-C t)^{2}},\) where \(B, C,\) and \(D\) are all positive constants. What is the speed of this wave?

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