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\(\bullet\) A certain transverse wave is described by the equation $$y(x, t)=(6.50 \mathrm{mm}) \sin 2 \pi\left(\frac{t}{0.0360 \mathrm{s}}-\frac{x}{0.280 \mathrm{m}}\right)$$ Determine this wave's (a) amplitude, (b) wavelength, (c) frequency, (d) speed of propagation, and (e) direction of propagation.

Short Answer

Expert verified
(a) Amplitude: 6.50 mm (b) Wavelength: 0.280 m (c) Frequency: 27.78 Hz (d) Speed: 7.78 m/s (e) Direction: Positive x-direction

Step by step solution

01

Identify the Amplitude

The amplitude of a wave is the coefficient of the sine function in the wave equation. In the given equation, \( y(x, t) = (6.50 \mathrm{mm}) \sin 2 \pi\left(\frac{t}{0.0360 \mathrm{s}} - \frac{x}{0.280 \mathrm{m}}\right) \), the amplitude is directly given as \( 6.50 \mathrm{mm} \).
02

Determine the Wavelength

The wavelength \( \lambda \) is determined from the spatial part of the wave number, which is the reciprocal of the coefficient of \( x \) inside the sine function, \( \frac{2\pi}{0.280 \mathrm{m}} \). Thus, the wavelength is \( \lambda = 0.280 \mathrm{m} \).
03

Calculate the Frequency

The frequency \( f \) can be found from the temporal part of the equation. Here, the term \( \frac{t}{0.0360 \mathrm{s}} \) gives the relationship between period \( T \) and frequency as \( T = 0.0360 \mathrm{s} \). Frequency is the inverse of the period, \( f = \frac{1}{T} = \frac{1}{0.0360 \mathrm{s}} \approx 27.78 \mathrm{Hz} \).
04

Find the Speed of Propagation

The speed of propagation \( v \) of the wave is given by the formula \( v = f \cdot \lambda \). Substituting the values calculated, \( v = 27.78 \mathrm{Hz} \times 0.280 \mathrm{m} = 7.78 \mathrm{m/s} \).
05

Determine the Direction of Propagation

The wave equation has a form \( \sin 2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right) \), indicating a positive relationship between time \( t \) and position \( x \). This suggests that 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.

Understanding Wave Amplitude
The concept of wave amplitude is central to understanding wave behavior. In any wave, the amplitude is the peak value of displacement from the wave's equilibrium position.
It's a measure of how "large" or "strong" a wave is. In the context of a transverse wave, like the one in our exercise, amplitude is the maximum distance the particles of the medium move from their rest position as the wave passes through.

In the given wave equation, the amplitude is the coefficient of the sine function. This means you can directly read off this value. For the wave described by the equation \(y(x, t) = (6.50\ \mathrm{mm}) \sin 2 \pi\left(\frac{t}{0.0360\ \mathrm{s}} - \frac{x}{0.280\ \mathrm{m}}\right)\), the amplitude is \(6.50\ \mathrm{mm}\). A larger amplitude indicates a wave that carries more energy.
Exploring Wavelength
Wavelength is a critical factor in wave phenomena. It is the distance between two consecutive points that are in phase, such as peak to peak or trough to trough.
In simpler terms, it's how "long" a single cycle of the wave is. This distance tells us about the spatial periodicity of the wave.

In the equation, the wavelength \(\lambda\) can be identified from the spatial component within the argument of the sine function. The wave number \(k\), defined as \(\frac{2\pi}{\lambda}\), has the coefficient \(\frac{1}{0.280\ \mathrm{m}}\). Therefore, the wavelength is \(\lambda = 0.280\ \mathrm{m}\). Longer wavelengths means the wave stretches out over a greater distance.
Frequency Uncovered
Frequency is directly related to how often the wave patterns repeat themselves over time. It tells us the number of complete cycles that pass a point in a unit of time,
often measured in Hertz (Hz), which corresponds to cycles per second. High frequency means more cycles occur per second.

For our transverse wave, frequency comes from the temporal part of the equation, given by the expression \(\frac{t}{0.0360\ \mathrm{s}}\). The period \(T\), the time for one complete cycle, is given as \(T = 0.0360\ \mathrm{s}\). The frequency \(f\) is calculated as the inverse of the period: \(f = \frac{1}{T} = \frac{1}{0.0360\ \mathrm{s}}\approx 27.78\ \mathrm{Hz}\). A higher frequency indicates a wave that oscillates more quickly.
Wave Speed Demystified
In wave mechanics, wave speed tells us how fast the wave front propagates through the medium.
It is determined by both the frequency and the wavelength of the wave. Essentially, it’s how quickly and in what time frame the pattern of the wave travels a certain distance.

Using the relationship \(v = f \cdot \lambda\), we calculate the speed for our specific wave. Substituting the frequency \(f\) of \(27.78\ \mathrm{Hz}\) and the wavelength \(\lambda\) of \(0.280\ \mathrm{m}\), the wave speed is \(v = 27.78\ \mathrm{Hz} \times 0.280\ \mathrm{m} = 7.78\ \mathrm{m/s}\). This means that in every second, the wave travels a distance of about 7.78 meters.
Direction of Wave Propagation
The direction of wave propagation is an essential characteristic, indicating which way the wave travels.
For transverse waves, this movement can be determined by the wave equation’s form.

Our wave equation is structured as \(\sin 2\pi\left(\frac{t}{T} - \frac{x}{\lambda}\right)\). This shows that the temporal and spatial terms have positive signs. A positive sign in front of the \(\frac{x}{\lambda}\) indicates the wave is moving in the positive x-direction.

In essence, this alignment suggests that as time increases, the wave pattern shifts to the right along the x-axis, confirming propagation towards increasing x-values.

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Most popular questions from this chapter

\(\bullet\) A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s for the boat to travel from its highest point to its lowest, a total distance of 0.62 \(\mathrm{m} .\) The fisherman sees that the wave crests are spaced 6.0 \(\mathrm{m}\) apart. (a) How fast are the waves traveling? (b) What is the amplitude of each wave? (c) If the total vertical distance traveled by the boat were \(0.30 \mathrm{m},\) but the other data remained the same, how would the answers to parts (a) and (b) be affected?

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