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Chapter 15: Problem 56

Transverse standing waves are produced on a string that has length \(0.800 \mathrm{~m}\) and is held fixed at each end. Each standing-wave pattern has a node at the fixed ends plus additional nodes along the length of the string. You measure the frequencies \(f_{n}\) for standing waves that have \(n\) of these nodes along their length. The tension in the string is kept constant. You plot \(n\) versus \(f_{n}\) and find that your data lie close to a straight line that has slope \(7.30 \times 10^{-3} \mathrm{~s}\). What is the speed of transverse waves on the string?

Short Answer

Expert verified
To find the speed of the wave, calculate it as twice the product of the given string's length and the line's slope from the plot (i.e., \(v = m \cdot 2L\)). Based on the provided values, the speed is \(v = 2 \times 7.3 \times 10^{-3} s^{-1} \times 0.800 m = 1.17 m/s\).

Step by step solution

01

Find the relationship between frequency and number of nodes

The relationship between the frequency of a standing wave and the number of nodes is given by the equation \(f_{n} = \frac{n \cdot v}{2L}\), where \(f_{n}\) is the frequency, \(n\) is the number of nodes, \(v\) is the wave speed, and \(L\) is the length of the string. When a plot of \(f_{n}\) vs \(n\) is straight line, we can write \(f_{n} = m \cdot n\) where \(m\) is the slope of the line.
02

Set up the equations for solution

We know from the problem that the slope of the line \(m = 7.30 \times 10^{-3} \mathrm{~s}\), and we also know that \(m = \frac{v}{2L}\) from our derived formula, since v and L are constants for a given string.
03

Solve for the wave speed

To find the wave speed, we equate our two formulas that include \(m\) and solve for \(v\), i.e., \(v = m \cdot 2L\). Using the given length \(L = 0.800 \mathrm{~m}\) and the calculated slope \(m = 7.30 \times 10^{-3} \mathrm{~s}\), we can compute the wave speed.

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Key Concepts

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

Wave Speed
Understanding wave speed is essential when it comes to transverse standing waves on a string. To put it simply, wave speed is the distance that the wave travels per unit of time, typically denoted as 'v'. It is determined by the medium through which the wave travels鈥攊n our case, the string鈥攁nd the tension applied to it. When dealing with a string, tension plays a significant role in how fast the wave moves: the greater the tension, the higher the wave speed and vice versa.

In the context of the problem provided, the wave speed is deduced from the relationship between the frequency of standing waves and the number of nodes, along with the length of the string. Since the tension in the string is held constant, the wave speed can be computed directly by multiplying the slope of the frequency versus nodes plot by two times the length of the string, symbolically represented as \( v = 2L \times m \). This provides a clear-cut method to calculate the speed at which waves propagate along the string.
Frequency of Standing Waves
The frequency of standing waves, often denoted as \(f_n\), is the number of oscillations that occur per second. In the context of standing waves on a string, each segment of the wave vibrates at the same frequency, which is intrinsically linked to the physical properties of the string, such as its length, tension, and mass.

The exercise poses an important relationship: the frequency of the standing wave is directly proportional to the number of nodes (points of no displacement). Mathematically, we examine this relationship through the formula \(f_{n} = \frac{n \cdot v}{2L}\). By analyzing the slope of the graph plotted with nodes and respective frequencies, we're provided with a direct way to find the missing wave speed. The frequency is integral to understanding the harmonics of the string鈥攖he series of frequencies at which the string naturally resonates, which is a concept deeply rooted in the physics of waves.
Nodes in Standing Waves
Nodes are quintessential features of standing waves, where they signify points of no motion along the medium鈥攊n this case, our string. When a standing wave is formed, certain points along the medium remain static as waves interfere with each other.

The location of these nodes is determined by the wave's frequency and the boundary conditions of the medium. For a string fixed at both ends, nodes appear at these fixed points and at integral multiples along the string's length. In simpler terms, the nodes divide the string into segments that vibrate in synchronization but in opposite phases. The number of these nodes is symbolically represented by 'n' in our calculations, and this count is vital when determining the frequency of the wave. It is worth noting that the patterns formed by nodes and antinodes (points of maximum displacement) help visualize the modal shapes of vibrations in the string鈥攁 beautiful demonstration of the wave's oscillatory nature.

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Most popular questions from this chapter

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^{7} \mathrm{~m}\). You suspend a lead weight from the lower end of a light string that is \(4.00 \mathrm{~m}\) long and has mass \(0.0280 \mathrm{~kg}\). You measure that it takes \(0.0685 \mathrm{~s}\) for a transverse pulse to travel from the lower end to the upper end of the string. On the earth, for the same string and lead weight, it takes \(0.0390 \mathrm{~s}\) for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

Standing waves are produced on a string that is held fixed at both ends. The tension in the string is kept constant. (a) For the second overtone standing wave the node-to-node distance is \(8.00 \mathrm{~cm} .\) What is the length of the string? (b) What is the node-to-node distance for the fourth harmonic standing wave?

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 (twice the frequency) than the note in part (a)?

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 transverse velocity of a particle \(1.35 \mathrm{~m}\) to the right of the origin at time \(t=0.0625 \mathrm{~s}\)

A sinusoidal transverse wave travels on a string. The string has length \(8.00 \mathrm{~m}\) and mass \(6.00 \mathrm{~g}\). The wave speed is \(30.0 \mathrm{~m} / \mathrm{s},\) and the wavelength is \(0.200 \mathrm{~m}\). (a) If the wave is to have an average power of \(50.0 \mathrm{~W}\), what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?
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