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A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antinode is \(1.80 \mathrm{~mm}\) and the speed of propagation of transverse waves on the string is \(260 \mathrm{~m} / \mathrm{s}\). The string extends along the \(x\) -axis, with one of the fixed ends at \(x=0,\) so that there is a node at \(x=0 .\) The smallest value of \(x\) where there is an antinode is \(x=0.150 \mathrm{~m}\). (a) What is the maximum transverse speed of a point on the string at an antinode? (b) What is the maximum transverse speed of a point on the string at \(x=0.075 \mathrm{~m} ?\)

Short Answer

Expert verified
The maximum transverse speed at an antinode is \(4.9 m/s\), and the maximum transverse speed at \(x = 0.075m\) is \(0 m/s\).

Step by step solution

01

Setup

Depending on whether a point on the string corresponds to a node or an antinode, its maximum transverse speed will vary. The maximum transverse speed at an antinode is given by the formula \(v_{max}= \omega A\), where \(\omega\) is the angular frequency of the wave and \(A\) is the amplitude of the wave. We know \(A = 1.80mm = 0.0018m\).
02

Calculate Angular Frequency

Because we know that there is an antinode at \(x=0.150m\), the distance between a node and an antinode is \(λ/4\). Therefore we have \(λ/4 = 0.150m\) which gives \(λ = 0.600m\). The speed of the waves is given by \(v = λf\), where \(f\) is frequency of the wave, and \(λ\) is wavelength. The speed is 260m/s and \(λ = 0.600m\), hence frequency \(f = v/λ = 260/0.6 = 433.33 Hz\). The angular frequency \(ω\) is given by \(2πf\), so \(ω = 2π ⋅ 433.33 = 2724.33 rad/s\).
03

Calculate Maximum Transverse speed at an Antinode

We can plug \(ω = 2724.33 rad/s\) and \(A = 0.0018m\) into \(v_{max} = ωA\) to get \(v_{max} = 2724.33 ⋅ 0.0018 = 4.9 m/s\).
04

Calculate Maximum Transverse Speed at Point \(x = 0.075m\)

The point at \(x = 0.075m\) is a node as it's half the distance between the node and the antinode. For standing waves, the transverse speed at a node is always zero. Thus, the maximum transverse speed at \(x = 0.075\) is \(0 m/s\).

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Key Concepts

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

Transverse Waves
Imagine a string fixed at both ends, and when you pluck it, waves that move perpendicular to the direction of the string's length are called transverse waves. The peaks and troughs you see moving along the string represent the high and low points of these waves.

Transverse waves consist of alternating high points, known as antinodes, where the wave reaches its maximum height, and low points, called nodes, where the movement is minimal or nonexistent. These nodes and antinodes are especially easy to spot in standing waves, which are waves that appear to be still even though they are the result of two identical waves moving in opposite directions, creating these fixed points of maximum and minimum displacement.
Angular Frequency
The angular frequency of a wave gives us a powerful way to describe its oscillations in terms of radians per second instead of the traditional cycles per second, or Hertz (Hz).

Employing the symbol \( \omega \) for angular frequency, it is related to the regular frequency \( f \) by the formula \( \omega = 2\pi f \). It tells us how many radians a transverse wave point rotates through in a second, which is particularly useful for waves and oscillations that are circular or rotational in nature.

This concept is crucial for understanding how points on a standing wave vibrate. The faster the angular frequency, the faster the points on the string move up and down.
Wavelength
The distance between consecutive identical points of a wave—like from peak to peak or from node to node—is known as wavelength, represented by the Greek letter \( \lambda \). It's a spatial period of the wave; the length over which the wave's shape repeats.

In the context of a standing wave, the wavelength can determine how nodes and antinodes are spaced along the medium. To be precise, the distance between a node and its neighbouring antinode is \( \lambda/4 \), because a full wavelength actually encompasses two nodes and two antinodes.

The wavelength is also directly tied to the speed \( v \) and frequency \( f \) of the wave through the relationship \( v = \lambda f \. \) Knowing the speed at which waves travel on a string and the frequency at which they oscillate allows us to calculate the wavelength precisely.

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Most popular questions from this chapter

You are designing a two-string instrument with metal strings \(35.0 \mathrm{~cm}\) long, as shown in Fig. \(\mathrm{P} 15.52 .\) Both strings are under the same tension. String \(S_{1}\) has a mass of \(8.00 \mathrm{~g}\) and produces the note middle \(\mathrm{C}\) (frequency \(262 \mathrm{~Hz}\) ) in its fundamental mode. (a) What should be the tension in the string? (b) What should be the mass of string \(S_{2}\) so that it will produce A-sharp (frequency \(466 \mathrm{~Hz}\) ) as its fundamental? (c) To extend the range of your instrument, you include a fret located just under the strings but not normally touching them. How far from the upper end should you put this fret so that when you press \(S_{1}\) tightly against it, this string will produce \(\mathrm{C}\) -sharp (frequency \(277 \mathrm{~Hz}\) ) in its fundamental? That is, what is \(x\) in the figure? (d) If you press \(S_{2}\) against the fret, what frequency of sound will it produce in its fundamental?

Transverse waves on a string have wave speed \(8.00 \mathrm{~m} / \mathrm{s},\) amplitude \(0.0700 \mathrm{~m},\) and wavelength \(0.320 \mathrm{~m} .\) The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a particle at \(x=0.360 \mathrm{~m}\) at time \(t=0.150 \mathrm{~s}\). (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{~m}\) next has maximum upward displacement?

A musician tunes the C-string of her instrument to a fundamental frequency of \(65.4 \mathrm{~Hz}\). The vibrating portion of the string is \(0.600 \mathrm{~m}\) long and has a mass of \(14.4 \mathrm{~g}\). (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from \(65.4 \mathrm{~Hz}\) to \(73.4 \mathrm{~Hz}\), corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D}\) ?

A \(1.50-\mathrm{m}\) -long rope is stretched between two supports with a tension that makes the speed of transverse waves \(62.0 \mathrm{~m} / \mathrm{s}\). What are the wavelength and frequency of (a) the fundamental; (b) the second overtone; (c) the fourth harmonic?

Standing waves on a wire are described by Eq. (15.28), with \(A_{\mathrm{SW}}=2.50 \mathrm{~mm}, \omega=942 \mathrm{rad} / \mathrm{s},\) and \(k=0.750 \pi \mathrm{rad} / \mathrm{m} .\) The left end of the wire is at \(x=0 .\) At what distances from the left end are (a) the nodes of the standing wave and (b) the antinodes of the standing wave?

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