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

Short Answer

Expert verified
(a) 6.50 mm; (b) 28.0 cm; (c) 27.78 Hz; (d) 778.84 cm/s; (e) positive x-direction.

Step by step solution

01

Identify the Amplitude

The amplitude of a wave is the maximum displacement from its equilibrium position. In the given wave equation \( y(x, t) = (6.50 \, \mathrm{mm}) \cos \left( 2\pi \left( \frac{x}{28.0 \, \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \right) \right) \), the amplitude is the coefficient in front of the cosine function. Here, it is \(6.50 \, \mathrm{mm}\).
02

Determine the Wavelength

The wavelength \( \lambda \) is given by the reciprocal of the coefficient of \(x\) inside the cosine function multiplied by \(2\pi\). Here, the coefficient is \( \frac{1}{28.0 \, \mathrm{cm}} \). Therefore, \( \lambda = 28.0 \, \mathrm{cm} \).
03

Calculate the Frequency

Frequency \( f \) is the reciprocal of the coefficient of \( t \) inside the cosine function multiplied by \(2\pi\). Here, the coefficient is \( \frac{1}{0.0360 \, \mathrm{s}} \). Thus, frequency is \( f = \frac{1}{0.0360 \, \mathrm{s}} = 27.78 \, \mathrm{Hz}\).
04

Calculate the Speed of Propagation

The speed \( v \) of the wave is calculated using the formula \( v = f \lambda \). Substituting the values from previous steps: \( v = 27.78 \, \mathrm{Hz} \times 28.0 \, \mathrm{cm} = 778.84 \, \mathrm{cm/s}\) or equivalently \(7.7884 \, \mathrm{m/s}\).
05

Determine the Direction of Propagation

The direction of wave propagation depends on the sign of the term in the cosine function. The term is \( \frac{x}{28.0 \, \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \), where \(x\) is positive and \(t\) is negative, indicating that the wave is traveling 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.

Amplitude
Amplitude is a fundamental property of waves, defining their maximum displacement from rest. In a wave, amplitude depicts how far the particles of the medium move from their equilibrium position due to the wave's energy. It essentially measures how 'strong' or 'intense' the wave is.

For the given wave equation, \(y(x, t) = (6.50 \, \mathrm{mm}) \cos \left( 2\pi \left( \frac{x}{28.0 \, \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \right) \right)\), the amplitude is represented by the coefficient in front of the cosine function. Hence, in this example, the amplitude is \(6.50 \, \mathrm{mm}\).

Amplitude is crucial for understanding wave energy; the greater the amplitude, the more energy the wave carries. It is one of the key parameters when analyzing wave behavior in physics and engineering.
Wavelength
Wavelength is the distance between consecutive identical points, like crests, in a wave pattern. It directly relates to the wave's spatial periodicity, defining the space over which the wave's shape repeats.

In our wave equation, the term \(\frac{x}{28.0 \, \mathrm{cm}}\) reveals the spatial component. The wavelength \(\lambda\) is found by taking the reciprocal of this coefficient, then multiplying by \(2\pi\). This simplifies directly as \(\lambda = 28.0 \, \mathrm{cm}\).

Wavelength plays a vital role in wave interactions and behaviors such as interference and diffraction. Understanding the wavelength helps in visualizing how waves propagate through various mediums.
Frequency
Frequency, denoted as \(f\), measures how often the wave oscillates in a unit of time, usually per second (Hertz, \(\mathrm{Hz}\)). It tells us how fast the wave cycles back and forth.

From the wave equation, the frequency is linked to the reciprocal of the coefficient of \(t\). In this case, it is \(\frac{1}{0.0360 \, \mathrm{s}}\), resulting in \(f = 27.78 \, \mathrm{Hz}\).

Frequency is a key characteristic in describing waves, particularly in contexts like sound, where pitch is related to frequency, or in electromagnetic waves where it relates to energy levels. Understanding frequency helps to quantify wave properties in scientific and practical applications.
Wave Speed
Wave speed is how fast a wave moves through a medium. It is calculated by the formula \(v = f \lambda\), meaning the wave speed is the product of frequency and wavelength. This relationship makes sense because wave speed combines how quickly the wave cycles (frequency) and how far each cycle travels (wavelength).

For this wave, substituting the derived values for frequency and wavelength gives us \(v = 27.78 \, \mathrm{Hz} \times 28.0 \, \mathrm{cm} = 778.84 \, \mathrm{cm/s}\), or \(7.7884 \, \mathrm{m/s}\).

Understanding wave speed allows us to predict how waves will move through different environments, a critical concept in engineering and physics fields like telecommunications and acoustics.
Wave Direction
The direction of wave propagation illustrates where the wave is traveling. By examining the terms in the wave equation, \( \frac{x}{28.0 \, \mathrm{cm}} - \frac{t}{0.0360 \, \mathrm{s}} \), the signs give insights into direction.

In this case, the term \( x \) is positive while \( t \) is negative, reflecting that the wave travels in the positive \(x\)-direction. This positive direction indicates that as time progresses, the points \( x \) where the wave reaches certain phases move positively along the \(x\)-axis.

Identifying the direction is vital for understanding wave behaviors and planning deployments in technologies such as radar, sonar, and wireless communication.

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Most popular questions from this chapter

A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspended from the lower end of the rope. What is the speed of transverse waves on the rope at the (a) bottom of the rope, (b) middle of the rope, and (c) top of the rope? (d) Is the tension in the middle of the rope the average of the tensions at the top and bottom of the rope? Is the wave speed at the middle of the rope the average of the wave speeds at the top and bottom? Explain.

One string of a certain musical instrument is 75.0 cm long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 m/s. (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 m? (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of 28.0 m/s. (a) What is the string's mass? (b) What is the magnitude of the maximum transverse acceleration of this point on the string?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

A 0.800-m-long string with linear mass density \(\mu = 7.50\) g/m is stretched between two supports. The string has tension \(F\) and a standing-wave pattern (not the fundamental) of frequency 624 Hz. With the same tension, the next higher standing-wave frequency is 780 Hz. (a) What are the frequency and wavelength of the fundamental standing wave for this string? (b) What is the value of \(F\)?

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