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These two waves travel along the same string: \(y_{1}(x, t)=(4.60 \mathrm{~mm}) \sin (2 \pi x-400 \pi t)\) \(y_{2}(x, t)=(5.60 \mathrm{~mm}) \sin (2 \pi x-400 \pi t+0.80 \pi \mathrm{rad}) .\) What are (a) the amplitude and (b) the phase angle (relative to wave 1 ) of the resultant wave? (c) If a third wave of amplitude \(5.00\) \(\mathrm{mm}\) is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

Short Answer

Expert verified
(a) 6.05 mm, (b) 1.25 rad, (c) 1.25 rad

Step by step solution

01

Identify and Write the Formula for Resultant Amplitude

The two given wave equations are in the form of \( y_1(x, t) = A_1 \sin(kx - \omega t) \) and \( y_2(x, t) = A_2 \sin(kx - \omega t + \phi) \). To find the resultant amplitude \( A_r \) when two waves are superposed, we use the formula: \[ A_r = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi)} \]Here, \( A_1 = 4.60 \text{ mm} \), \( A_2 = 5.60 \text{ mm} \), and \( \phi = 0.80 \pi \text{ rad} \).
02

Calculate Resultant Amplitude

Substitute the values into the formula:\[ A_r = \sqrt{(4.60)^2 + (5.60)^2 + 2 \times 4.60 \times 5.60 \times \cos(0.80\pi)} \]\[ A_r = \sqrt{21.16 + 31.36 + 51.52 \times \cos(0.80\pi)} \]Calculating \( \cos(0.80\pi) = -0.309 \), we find:\[ A_r = \sqrt{21.16 + 31.36 - 15.91} \]\[ A_r = \sqrt{36.61} \approx 6.05 \text{ mm} \]
03

Determine the Phase Angle of the Resultant Wave

The phase angle of the resultant wave \( \phi_r \) can be calculated using:\[ \tan(\phi_r) = \frac{A_2 \sin(\phi)}{A_1 + A_2 \cos(\phi)} \]Substitute the values:\[ \tan(\phi_r) = \frac{5.60 \sin(0.80\pi)}{4.60 + 5.60 \cos(0.80\pi)} \]Here \( \sin(0.80\pi) = 0.951 \) and \( \cos(0.80\pi) = -0.309 \):\[ \tan(\phi_r) = \frac{5.32}{1.84} = 2.891 \]Thus, \( \phi_r = \tan^{-1}(2.891) \approx 71.6^\circ \text{ or } 1.25 \text{ rad} \).
04

Determine Phase Angle for Third Wave

To maximize the resultant amplitude, the third wave \( y_3 = A_3 \sin(kx - \omega t + \phi_3) \) should be in phase with the resultant wave of the two previous waves. Thus, the phase angle \( \phi_3 \) of the third wave should be made equal to the resultant phase angle \( \phi_r \) to ensure constructive interference. Thus, \( \phi_3 = 1.25 \text{ rad} \).
05

Result Summary

To summarize: - The amplitude of the resultant wave is \( 6.05 \text{ mm} \).- The phase angle relative to wave 1 is \( 1.25 \text{ rad} \) or approximately \( 71.6^\circ \).- The phase angle for the third wave to maximize amplitude should also be \( 1.25 \text{ rad} \).

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Key Concepts

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

Amplitude Calculation
When two waves travel along the same path, they interfere with each other, which can result in a new, combined wave known as the *resultant wave*. A key component of understanding wave interference is calculating the amplitude of this resultant wave.
To find this amplitude, we use the formula:
  • \[ A_r = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\phi)} \]
Here, \( A_1 \) and \( A_2 \) are the amplitudes of the two initial waves, and \( \phi \) is the phase difference between them. This equation takes into account both the individual amplitudes and the phase difference, allowing us to understand how much they will amplify or diminish each other's effects.
Using the values from the example, with \( A_1 = 4.60 \text{ mm} \), \( A_2 = 5.60 \text{ mm} \), and \( \phi = 0.80 \pi \text{ rad} \), you can calculate:
  • \[ A_r = \sqrt{(4.60)^2 + (5.60)^2 + 2 \times 4.60 \times 5.60 \times \cos(0.80\pi)} \]
Knowing \( \cos(0.80\pi) = -0.309 \), we find the resultant amplitude to be approximately \( 6.05 \text{ mm} \). This shows how the initial waves combine to create a single wave with greater amplitude because of both constructive and destructive interference.
Phase Angle
The phase angle is crucial in understanding how two waves combine. It describes the difference in the wave positions at a given point in time, relative to each other. This phase difference can determine whether the waves interfere constructively (amplifying the resultant wave) or destructively (diminishing it).
To calculate the resultant wave's phase angle \( \phi_r \) relative to the first wave, we apply:
  • \[ \tan(\phi_r) = \frac{A_2 \sin(\phi)}{A_1 + A_2 \cos(\phi)} \]
In our example, this involves substituting \( A_1 = 4.60 \text{ mm} \), \( A_2 = 5.60 \text{ mm} \), \( \sin(0.80\pi) = 0.951 \), and \( \cos(0.80\pi) = -0.309 \):
  • \[ \tan(\phi_r) = \frac{5.60 \times 0.951}{4.60 - 5.60 \times 0.309} \]
  • \[ \tan(\phi_r) = \frac{5.32}{1.84} = 2.891 \]
The resultant phase angle, \( \phi_r \), can then be calculated using the inverse tangent function, yielding approximately \( 71.6^\circ \) or \( 1.25 \text{ rad} \).
Understanding phase angles helps visualize how wave phases align or misalign, leading to variations in amplitude.
Resultant Wave
When discussing wave interference, the *resultant wave* is the new wave formed by the combination of two or more waves traveling along the same medium. The attributes of this resultant wave, such as its amplitude and phase angle, indicate how the original waves interact.
For maximum amplitude in wave interference, the waves should interfere constructively. We achieve this by adjusting the phase angles of subsequent waves to match the resultant wave's phase angle. In our example, this means adding a third wave with an appropriate phase adjustment. To do this, the phase angle of the third wave \( \phi_3 \) should match the resultant phase angle, previously calculated as \( 1.25 \text{ rad} \).
  • When the resultant wave's attributes are aligned perfectly with any additional waves, this ensures maximum amplitude due to constructive interference.
  • This constructive interference is the principle behind technologies like noise-canceling headphones and radio broadcasting, where wave manipulation is crucial.
Thus, understanding how to develop and manipulate resultant waves allows for practical applications in various fields of science and engineering.

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Most popular questions from this chapter

Two sinusoidal waves of the same wavelength travel in the same direction along a stretched string. For wave \(1, y_{m}=3.0 \mathrm{~mm}\) and \(\phi=\) \(0 ;\) for wave \(2, y_{m}=5.0 \mathrm{~mm}\) and \(\phi=70^{\circ} .\) What are the (a) amplitude and (b) phase constant of the resultant wave?

The function \(y(x, t)=(15,0 \mathrm{~cm}) \cos (\pi x-15 \pi t)\), with \(x\) in meters and \(t\) in seconds, describes a wave on a taut string. What is the transverse speed for a point on the string at an instant when that point has the displacement \(y=+12.0 \mathrm{~cm} ?\)

The equation of a transverse wave on a string is $$ y=(2.0 \mathrm{~mm}) \sin \left[\left(20 \mathrm{~m}^{-1}\right) x-\left(600 \mathrm{~s}^{-1}\right) t\right] $$ The tension in the string is \(15 \mathrm{~N}\). (a) What is the wave speed? (b) Find the linear density of this string in grams per meter.

A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (Fig. \(16-29\) ). The waves are of two types: transverse waves traveling at \(v_{t}=50 \mathrm{~m} / \mathrm{s}\) and longitudinal waves traveling at \(v_{l}=150 \mathrm{~m} / \mathrm{s} .\) If a sudden motion sends out such waves, a scorpion can tell the distance of the beetle from the difference \(\Delta t\) in the arrival times of the waves at its leg nearest the beetle. If \(\Delta t=4.0 \mathrm{~ms}\) what is the beetle's distance?

A sinusoidal transverse wave of amplitude \(y_{m}\) and wavelength \(\lambda\) travels on a stretched cord. (a) Find the ratio of the maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed. (b) Does this ratio depend on the material of which the cord is made?

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