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(II) A transverse traveling wave on a cord is represented by \(D=0.22 \sin (5.6 x+34 t)\) where \(D\) and \(x\) are in meters and \(t\) is in seconds. For this wave determine (a) the wavelength, (b) frequency, (c) velocity (magnitude and direction), (d) amplitude, and (e) maximum and minimum speeds of particles of the cord.

Short Answer

Expert verified
(a) 1.12 m, (b) 5.41 Hz, (c) 6.06 m/s; negative direction, (d) 0.22 m, (e) Max speed: 7.48 m/s, Min speed: 0 m/s.

Step by step solution

01

Identify the Wave Equation

The given wave equation is\[ D = 0.22 \sin (5.6x + 34t) \]where \(D\) is the displacement, \(x\) is the position, and \(t\) is the time. The equation is in the standard form \( D = A \sin(kx + \omega t) \).
02

Determine the Amplitude

From the wave equation \( D = 0.22 \sin (5.6x + 34t) \), the amplitude \( A \) is the coefficient in front of the sine function. Thus, the amplitude is \( A = 0.22 \text{ meters} \).
03

Calculate the Wavelength

In the equation \( D = A \sin(kx + \omega t) \), \(k\) is the wave number, which is equal to \( 5.6 \). The wavelength \( \lambda \) is given by the formula \( \lambda = \frac{2\pi}{k} \). Substituting \( k = 5.6 \), we have:\[ \lambda = \frac{2\pi}{5.6} \approx 1.12 \text{ meters} \]
04

Determine the Frequency

The angular frequency \( \omega \) is given as \( 34 \). The frequency \( f \) can be found using the relation \( \omega = 2\pi f \). Solving for \( f \), we have:\[ f = \frac{\omega}{2\pi} = \frac{34}{2\pi} \approx 5.41 \text{ Hz} \]
05

Calculate the Wave Velocity

The wave velocity \( v \) can be determined using the equation \( v = f\lambda \). With \( f = 5.41 \text{ Hz} \) and \( \lambda = 1.12 \text{ meters} \), we find:\[ v = 5.41 \times 1.12 \approx 6.06 \text{ m/s} \]The direction of wave propagation is determined by the sign in front of \(kx\) and \(\omega t\); since they are positive, the wave propagates in the negative \(x\)-direction.
06

Determine Maximum Particle Speed

The maximum speed of particles on the cord occurs when the derivative of displacement with respect to time is maximized. The speed of particles is given by \( v_{particle} = A \omega \cos(kx + \omega t) \). The maximum value of \( \cos \) is 1, therefore:\[ v_{particle, max} = A \omega = 0.22 \times 34 = 7.48 \text{ m/s} \]
07

Determine Minimum Particle Speed

The minimum speed is 0 m/s, which occurs when the particle is at the equilibrium position (midpoint of oscillation). This happens when \( \cos(kx + \omega t) = 0 \).

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Key Concepts

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

Wavelength Calculation
Understanding how to calculate the wavelength of a wave is crucial in wave mechanics. The wavelength (\( \lambda \)) represents the distance between two consecutive points that are in phase, such as the distance between two peaks or troughs of a wave.
In our problem, we have a wave described by the equation:\[D = 0.22 \sin(5.6x + 34t)\]
The wave number (\( k \)) is given as 5.6. We can find the wavelength using the formula:\[\lambda = \frac{2\pi}{k}\]
Substituting\( k = 5.6 \), we get:\[\lambda = \frac{2\pi}{5.6} \approx 1.12 \text{ meters}\]
Thus, the wavelength of this wave is approximately 1.12 meters.
Frequency Determination
Frequency refers to how many cycles or oscillations a wave completes in one second. It is measured in Hertz (Hz). Knowing the frequency helps us understand how often the wave peaks.
In the given equation:\[D = 0.22 \sin(5.6x + 34t)\]
The angular frequency (\( \omega \)) is provided as 34. We can calculate the frequency (\( f \)) using the relationship:\[\omega = 2\pi f\]
Rearranging for\( f \), gives:\[f = \frac{\omega}{2\pi} = \frac{34}{2\pi} \approx 5.41 \text{ Hz}\]
This tells us that the wave completes approximately 5.41 cycles per second.
Wave Velocity
Wave velocity is the speed and direction in which the wave travels through the medium. It can be determined from the frequency and wavelength.
The wave velocity (\( v \)) is calculated using the equation:\[v = f\lambda\]
With\( f = 5.41 \text{ Hz} \) and\( \lambda = 1.12 \text{ meters} \), we find:\[v = 5.41 \times 1.12 \approx 6.06 \text{ m/s}\]
This means the wave travels at a speed of approximately 6.06 meters per second. The direction of travel is determined from the equation whether the terms involving\( x \)t and\( t \) are positive or negative. In this case, since both\( 5.6x \) and\( 34t \) are positive, the wave moves in the negative x-direction.
Amplitude
Amplitude is the maximum displacement of the particles from their equilibrium position in the medium through which the wave travels. It gives us an idea of the wave's energy.
In our wave equation:\[D = 0.22 \sin(5.6x + 34t)\]
The amplitude (\( A \)) is represented by the coefficient before the sine function, which is 0.22.
This means the particles move 0.22 meters from their rest position in both directions.
  • The larger the amplitude, the more energy the wave carries.
  • In practical terms, amplitude affects the loudness of sound waves and brightness of light waves.
Particle Speed in Waves
Particles in the medium through which the wave travels oscillate back and forth. Their speed varies over time during this motion.
In the context of our wave, the particle speed (\( v_{\text{particle}} \)) can be calculated by differentiating the displacement with respect to time:\[v_{\text{particle}} = A \omega \cos(kx + \omega t)\]
Given\( A = 0.22 \) and\( \omega = 34 \), the maximum particle speed occurs when\( \cos \) is 1, so:\[v_{\text{particle, max}} = 0.22 \times 34 = 7.48 \text{ m/s}\]
This maximum value indicates how fast particles can move in the medium.
The minimum speed is 0 m/s and occurs when the particle reaches its equilibrium position, where there is no displacement.

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Most popular questions from this chapter

(II) A guitar string is 90.0 \(\mathrm{cm}\) long and has a mass of 3.16 \(\mathrm{g}\) . From the bridge to the support post \((=\ell)\) is 60.0 \(\mathrm{cm}\) and the string is under a tension of 520 \(\mathrm{N}\) . What are the frequencies of the fundamental and first two overtones?

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(II) If two successive harmonics of a vibrating string are \(240 \mathrm{~Hz}\) and \(320 \mathrm{~Hz},\) what is the frequency of the fundamental?

(I) Water waves approach an underwater "shelf" where the velocity changes from \(2.8 \mathrm{~m} / \mathrm{s}\) to \(2.5 \mathrm{~m} / \mathrm{s}\). If the incident wave crests make a \(35^{\circ}\) angle with the shelf, what will be the angle of refraction?

(II) Suppose two linear waves of equal amplitude and frequency have a phase difference \(\phi\) as they travel in the same medium. They can be represented by $$ \begin{array}{l} D_{1}=A \sin (k x-\omega t) \\ D_{2}=A \sin (k x-\omega t+\phi) \end{array} $$ \(\begin{array}{llll}\text { (a) Use the trigonometric identity } & \sin \theta_{1}+\sin \theta_{2}=\end{array}\) \(2 \sin \frac{1}{2}\left(\theta_{1}+\theta_{2}\right) \cos \frac{1}{2}\left(\theta_{1}-\theta_{2}\right)\) to show that the resultant wave is given by $$ D=\left(2 A \cos \frac{\phi}{2}\right) \sin \left(k x-\omega t+\frac{\phi}{2}\right) $$ (b) What is the amplitude of this resultant wave? Is the wave purely sinusoidal, or not? (c) Show that constructive interference occurs if \(\phi=0,2 \pi, 4 \pi,\) and so on, and destructive interference occurs if \(\phi=\pi, 3 \pi, 5 \pi,\) etc. \((d)\) Describe the resultant wave, by equation and in words, if \(\phi=\pi / 2\).

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