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Chapter 12: Problem 9

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
Amplitude: 6.50 mm, Wavelength: 0.280 m, Frequency: 27.78 Hz, Speed: 7.78 m/s, Direction: positive x-direction.

Step by step solution

01

Identify the Amplitude

The standard equation for a transverse wave is given by:\[y(x, t) = A \sin(2 \pi(ft - \frac{x}{\lambda}))\]where \(A\) is the amplitude. In our 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 \(A\) is \(6.50 \, \mathrm{mm}\).
02

Calculate the Wavelength

Comparing the given wave equation with the standard wave form, the term \(\frac{x}{\lambda}\) corresponds to \(\frac{x}{0.280}\). Hence, the wavelength \(\lambda = 0.280 \, \mathrm{m}\).
03

Determine the Frequency

The term \(\frac{t}{T}\) corresponds to \(\frac{t}{0.0360 \, \mathrm{s}}\), where \(T\) is the period of the wave. Hence, \(T = 0.0360 \, \mathrm{s}\). Frequency \(f\) is the reciprocal of the period:\[f = \frac{1}{T} = \frac{1}{0.0360} \, \mathrm{Hz} \approx 27.78 \, \mathrm{Hz}.\]
04

Calculate the Speed of Propagation

The speed of the wave \(v\) is given by the product of frequency and wavelength:\[v = f \times \lambda = 27.78 \, \mathrm{Hz} \times 0.280 \, \mathrm{m} \approx 7.78 \, \mathrm{m/s}.\]
05

Determine the Direction of Propagation

The direction of propagation is determined by the sign of \(kx - \omega t\) in the wave equation. Since the term is \(\frac{t}{0.0360 \, \mathrm{s}} - \frac{x}{0.280 \, \mathrm{m}}\), it indicates the wave is propagating 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 key characteristic of a wave. It describes the height of the wave crest or depth of the trough. Think of amplitude as the "loudness" of the wave. For a wave traveling through water, it's how high each wave gets. In our wave equation:\[y(x, t)=(6.50 \, \mathrm{mm}) \sin\left(2 \pi\left(\frac{t}{0.0360 \, \mathrm{s}}-\frac{x}{0.280 \, \mathrm{m}}\right)\right)\]The amplitude \(A\) is given by the term in front of the sine function. Here, it is 6.50 mm.
  • The amplitude measures the wave's maximum displacement from its equilibrium position.
  • Physical contexts like sound use bigger amplitudes for louder sounds.
Understanding amplitude helps comprehend the energy a wave carries. The larger the amplitude, the more energy is transferred by the wave.
Wavelength
Wavelength is the distance over which the wave's shape repeats. In simple terms, it's the distance from one crest of the wave to the next. Mathematically, the wavelength \(\lambda\) is found in the wave equation as:\[y(x, t)= (6.50 \, \mathrm{mm}) \sin\left(2 \pi\left(\frac{t}{0.0360 \, \mathrm{s}}-\frac{x}{0.280 \, \mathrm{m}}\right)\right)\]The wavelength can be deduced from the comparison to the standard form of the wave equation. We identify that:\[ \frac{x}{\lambda} = \frac{x}{0.280}\]This implies that \(\lambda = 0.280 \, \mathrm{m}\).These crests and troughs measure how spaced out the wave repetitions are.
Wave Frequency
Wave frequency defines how often the wave's crests pass a particular point per second. In the context of sound waves, frequency determines the pitch of the sound. For our wave, frequency \(f\) is the reciprocal of the period \(T\), and is derived from:\[\frac{t}{T} = \frac{t}{0.0360 \, \mathrm{s}}\]\[\therefore \, T = 0.0360 \, \mathrm{s}\]Thus, frequency \(f\) is:\[f = \frac{1}{T} = \frac{1}{0.0360 \, \mathrm{s}} \approx 27.78 \, \mathrm{Hz}\]
  • More cycles per second mean higher frequency.
  • Frequency is measured in Hertz (Hz), which means cycles per second.
Higher frequencies result in sound waves with higher pitches.
Wave Propagation
Wave propagation refers to how a wave travels through a medium. It's essential to know the speed, which indicates how fast a wave front moves.The speed \(v\) of our wave is calculated from the equation:\[v = f \times \lambda = 27.78 \, \mathrm{Hz} \times 0.280 \, \mathrm{m} \approx 7.78 \, \mathrm{m/s}\]
  • The wave speed is derived from both the frequency and the wavelength.
  • It's like a car driving continuously along the crest of a wave in time.
Understanding wave speed is crucial for predicting when a wave will reach a certain point.
Wave Direction
Wave direction describes the path along which the wave energy is traveling. In our 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 sign of the expression within the sine function helps identify the direction. Since it's "\(\frac{t}{0.0360 \, \mathrm{s}}-\frac{x}{0.280 \, \mathrm{m}}\)", this means the wave propagates in the positive x-direction.
  • Positive signs for \(x\) imply movement in the positive direction.
  • Understanding direction helps in mapping where and when to expect wave effects.
Recognizing the wave's direction is vital in applications like predicting the arrival of ocean waves on a shoreline.

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Most popular questions from this chapter

Transverse waves on a string have wave speed \(8.00 \mathrm{~m} / \mathrm{s}\), amplitude \(0.0700 \mathrm{~m},\) and wavelength \(0.320 \mathrm{~m}\). These waves travel in the \(x\) direction, and at \(t=0\) the \(x=0\) end of the string is at \(y=0\) and moving downward. (a) Find the frequency, period, and wave number \(k=2 \pi / \lambda\) of these waves. (b) Write the equation for \(y(x, t)\) describing these waves. (c) Find the transverse displacement of a point on the string at \(x=0.360 \mathrm{~m}\) at time \(t=0.150 \mathrm{~s}\)

\(\mathrm{A} 75.0 \mathrm{~cm}\) wire of mass \(5.625 \mathrm{~g}\) is tied at both ends and adjusted to a tension of \(35.0 \mathrm{~N}\). When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

Two train whistles, \(A\) and \(B\), each have a frequency of \(392 \mathrm{~Hz}\). \(A\) is stationary and \(B\) is moving toward the right (away from \(A\) ) at a speed of \(35.0 \mathrm{~m} / \mathrm{s}\). A listener is between the two whistles and is moving toward the right with a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). (See Figure \(12.43 .)\) (a) What is the frequency from \(A\) as heard by the listener? (b) What is the frequency from \(B\) as heard by the listener? (c) What is the beat frequency detected by the listener?

The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length \(L\). what fraction of \(L\) should be cut off so that the fundamental mode of the resulting open pipe has the same frequency as the fifth harmonic \((n=5)\) of the original stopped pipe?

A railroad train is traveling at \(30.0 \mathrm{~m} / \mathrm{s}\) in still air. The frequency of the note emitted by the train whistle is \(262 \mathrm{~Hz}\). What frequency is heard by a passenger on a train moving in the opposite direction to the first at \(18.0 \mathrm{~m} / \mathrm{s}\) and (a) approaching the first; and (b) receding from the first?
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