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Chapter 15: Problem 8

A certain transverse wave is described by $$ y(x, t)=(6.50 \mathrm{~mm}) \cos 2 \pi\left(\frac{x}{28.0 \mathrm{~cm}}-\frac{t}{0.0360 \mathrm{~s}}\right) $$ Determine the wave's (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

Short Answer

Expert verified
The wave's (a) amplitude is \(6.50 mm\), (b) wavelength is \(28.0 cm\), (c) frequency is \(27.8 Hz\), (d) speed of propagation is \(778 cm/s\), and (e) direction of propagation is in the positive x direction.

Step by step solution

01

Identify the Amplitude

The amplitude of a wave is the maximum displacement of the wave. It is the coefficient of the cosine function. Therefore, in this case, the amplitude of the wave is the coefficient, \(6.50 mm\). This is already in the correct unit, so no further conversion is needed.
02

Identify the Wavelength

The wavelength of a wave is the distance over which the wave's shape is repeated, and it is \(\frac{2\pi}{\text{coefficient of } x}\) for this cosine wave. Therefore, in this case, the wavelength is \(\frac{2\pi}{2\pi / 28.0 cm}\), which simplifies to \(28.0 cm\). This is already in the correct unit, so no further conversion is needed.
03

Identify the Frequency

The frequency of a wave is the number of complete wave cycles per unit time. It is \(\frac{2\pi}{\text{coefficient of } t}\) for this cosine wave. In this case, the frequency is \(\frac{2\pi}{2\pi / 0.0360 s}\), which simplifies to \(1 / 0.0360 s^{-1}\) or approximately \(27.8 Hz\). Note that \(s^{-1}\) is equivalent to the unit Hertz (Hz).
04

Identify the Speed of Propagation

The speed of propagation of a wave can be found by multiplying the wavelength by the frequency. In this case, \(28.0 cm \times 27.8 Hz = 778 cm/s\). Note that \(cm/s\) is a unit of speed.
05

Identify the Direction of Propagation

The direction of propagation determines whether the wave moves in the positive or negative direction along the x-axis. Because the cosine function has a negative sign in front of the time variable, the wave propagates in the positive x direction.

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Key Concepts

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

Wave Amplitude
Wave amplitude is a fundamental characteristic of waves that relates to the peak height from the wave's rest position. In simpler terms, it's how tall the wave peaks are, or how deep the troughs sink. The higher the amplitude, the more energy the wave carries. In our exercise, the amplitude is given by the coefficient in front of the cosine function, yielding an amplitude of \(6.50 \text{mm}\). Just imagine a calm sea with small ripples versus large, crashing waves during a storm 鈥� the storm waves have much higher amplitudes.
Wave Wavelength
Wavelength is the spatial period of a wave, meaning the distance over which the wave's shape repeats. It is often seen as the distance between consecutive crests or troughs. Think of the wavelength as the length of one cycle of the wave. The wavelength can be found from the wave equation and in our example, the wavelength corresponds to \(28.0 \text{cm}\). In visual terms, wavelength is how long or short the waves are from one peak to the next.
Wave Frequency
Frequency is how often the waves pass by a point in a given time and is measured in cycles per second, or Hertz (Hz). It tells you how fast the wave oscillates. In terms of our example, the frequency comes out to be around \(27.8 \text{Hz}\), meaning each second, about 28 cycles of the wave would pass a fixed point. The higher the frequency, the more waves you get in a set amount of time.
Speed of Wave Propagation
The speed of wave propagation is the rate at which the wave travels through the medium. It's calculated by multiplying the wavelength by the frequency. From the exercise, we've identified the wave's speed to be \(778 \text{cm/s}\), suggesting that the wave moves quite fast, covering about 7.78 meters every second! The speed tells you how quickly the energy of the wave is being transmitted.
Direction of Wave Propagation
Finally, the direction of wave propagation is the way in which the wave travels through space. For transverse waves, like on a string, this would be at right angles to the motion of the wave energy. In the case of our exercise, the negative sign in front of the time variable in the cosine function implies that this wave is moving in the positive x-direction. That means if you were to sit and watch this wave, you would see the peaks and troughs moving from left to right, much like watching ocean waves come onto shore.

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Most popular questions from this chapter

A piano wire with mass \(3.00 \mathrm{~g}\) and length \(80.0 \mathrm{~cm}\) is stretched with a tension of \(25.0 \mathrm{~N}\). A wave with frequency \(120.0 \mathrm{~Hz}\) and amplitude \(1.6 \mathrm{~mm}\) travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

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?

For a violin, estimate the length of the portions of the strings that are free to vibrate. (a) The frequency of the note played by the open E5 string vibrating in its fundamental standing wave is 659 Hz. Use your estimate of the length to calculate the wave speed for the transverse waves on the string. (b) The vibrating string produces sound waves in air with the same frequency as that of the string. Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and calculate the wavelength of the E5 note in air. Which is larger: the wavelength on the string or the wavelength in air? (c) Repeat parts (a) and (b) for a bass viol, which is typically played by a person standing up. Start your calculation by estimating the length of the bass viol string that is free to vibrate. The G2 string produces a note with frequency \(98 \mathrm{~Hz}\) when vibrating in its fundamental standing wave.

A horizontal wire is tied to supports at each end and vibrates in its second- overtone standing wave. The tension in the wire is \(5.00 \mathrm{~N}\), and the node-to-node distance in the standing wave is \(6.28 \mathrm{~cm}\). (a) What is the length of the wire? (b) A point at an antinode of the standing wave on the wire travels from its maximum upward displacement to its maximum downward displacement in \(8.40 \mathrm{~ms}\). What is the wire's 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?
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