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These two waves travel along the same string: $$ \begin{aligned} &y_{1}(x, t)=(4.00 \mathrm{~mm}) \sin (2 \pi x-650 \pi t) \\ &y_{2}(x, t)=(6.20 \mathrm{~mm}) \sin (2 \pi x-650 \pi t+0.60 \pi \mathrm{rad}) \end{aligned} $$ 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) Amplitude is 6.26 mm, (b) Phase angle is 70.83° relative to wave 1. (c) Third wave should have the same phase angle of 70.83°.

Step by step solution

01

Understand the Wave Equations

We are given two sinusoidal waves with the following equations:\[ y_1(x, t) = 4.00 \text{ mm} \sin(2 \pi x - 650 \pi t) \]\[ y_2(x, t) = 6.20 \text{ mm} \sin(2 \pi x - 650 \pi t + 0.60 \pi) \]These waves share the same angular wave number and frequency, suggesting they are in phase-related conditions.
02

Use the Superposition Principle

Since the waves travel along the same string, the principle of superposition states that the resultant wave is the sum of the individual waves:\[ y(x, t) = y_1(x, t) + y_2(x, t) \]
03

Find the Resultant Amplitude

The amplitude of the resultant wave can be found using vectors, as the resultant amplitude is the vector sum of two wave amplitudes:\[ A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi} \]where \( A_1 = 4.00 \text{ mm} \), \( A_2 = 6.20 \text{ mm} \), and \( \phi = 0.60 \pi \):\[ A = \sqrt{4.00^2 + 6.20^2 + 2 \times 4.00 \times 6.20 \times \cos(0.60 \pi)} \]
04

Calculate \( \cos(0.60 \pi) \)

Solving the above gives:\[ \cos(0.60 \pi) = \cos(108^\circ) = -0.309 \]
05

Calculate Amplitude

Substitute \( \cos(0.60 \pi) = -0.309 \) into the equation for amplitude:\[ A = \sqrt{16 + 38.44 - 15.288} = \sqrt{39.152} = 6.26 \text{ mm} \].
06

Calculate Resultant Phase Angle

The resultant phase angle \( \theta \) is given by:\[ \tan \theta = \frac{A_2 \sin \phi}{A_1 + A_2 \cos \phi} \]Substitute \( A_1 = 4.00 \), \( A_2 = 6.20 \), and \( \phi = 0.60 \pi \):\[ \tan \theta = \frac{6.20 \times \sin(0.60 \pi)}{4.00 + 6.20 \times (-0.309)} \]
07

Calculate \( \sin(0.60 \pi) \)

\[ \sin(0.60 \pi) = \sin(108^\circ) = 0.951 \]
08

Evaluate \( \tan \theta \) and \(\theta\)

Substituting the sine value:\[ \tan \theta = \frac{6.20 \times 0.951}{4.00 - 6.20 \times 0.309} = \frac{5.8962}{2.0878} \approx 2.823 \]Thus:\[ \theta = \arctan(2.823) \approx 70.83^\circ \].
09

Introduce Third Wave to Maximize Amplitude

To maximize the amplitude by introducing a third wave with amplitude 5.00 mm, its phase angle should be \( \theta_3 = \theta = 70.83^\circ \). This ensures constructive interference resulting in maximal amplitude.

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Key Concepts

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

Superposition Principle
The Superposition Principle is a fundamental concept in wave mechanics. It states that when two or more waves traverse the same space, their displacements add together algebraically at every point. This principle helps us understand complex wave interactions and solve real-world problems by breaking them into simpler components.
For instance, when waves travel along the same string, the resultant wave can be accurately predicted by adding the individual waves' equations. It’s like combining overlapping pictures to create a composite image. In our exercise, the two given waves are described by the equations \( y_1(x, t) \) and \( y_2(x, t) \).
Their resultant wave, \( y(x, t) \), is created by adding these two:
  • \( y(x, t) = y_1(x, t) + y_2(x, t) \)
This concept explains how waves interfere with each other, leading to either an increase or decrease in the overall amplitude depending on their relative phases.
Resultant Amplitude
Resultant Amplitude is the combined amplitude of multiple waves, measuring the wave's maximum displacement. Calculating this involves taking into account both amplitude magnitudes and their relative phase angles.
To solve for the resultant amplitude, the formula used is:
  • \( A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi} \)
Here, \( A_1 \) and \( A_2 \) represent individual wave amplitudes, and \( \phi \) is the phase difference. In our exercise, these values are plugged in to obtain:
  • \( A_1 = 4.00 \text{ mm} \)
  • \( A_2 = 6.20 \text{ mm} \)
  • \( \phi = 0.60 \pi \)
This results in a calculated amplitude of about 6.26 mm, combining the influences of both contributing waves.
Phase Angle
The Phase Angle is a critical concept in understanding how waves interfere with each other. It represents the offset between wave cycles and governs the wave interference pattern.
To determine the resultant phase angle, we consider the phase angles of individual waves and their interactions, which can be calculated using the formula:
  • \( \tan \theta = \frac{A_2 \sin \phi}{A_1 + A_2 \cos \phi} \)
For the given exercise, after substituting values:
  • \( \tan \theta = \frac{6.20 \times 0.951}{4.00 + 6.20 \times (-0.309)} \)
This calculates the resultant phase angle \( \theta \) as approximately 70.83 degrees. This angle is pivotal for predicting how the wave behaves under interference conditions.
Constructive Interference
Constructive Interference occurs when two or more waves align such that their crests and troughs coincide, resulting in a higher amplitude. This is the 'adding up' of waves that leads to an enhanced wave effect.
Maximizing amplitude through constructive interference involves tuning phase angles to ensure wave peaks line up. With a third wave introduced to the scenario described in the exercise, it can achieve maximized amplitude—a key outcome of constructive interference—by setting its phase angle to align with the resultant wave phase angle.
  • This means setting its phase angle to approximately \( 70.83^{\circ} \) to match the previously calculated angle for maximum collective amplitude.
Thus, by aligning wave phases, we exploit constructive interference to amplify the resulting wave efficiently.

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Most popular questions from this chapter

A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(6.00 \mathrm{~s}^{-1}\right) t\right] $$ and a generator at the other end creates the wave $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x-\left(6.00 \mathrm{~s}^{-1}\right) t\right] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For \(x \geq 0\), what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of \(x\) ? For \(x \geq 0\), what is the location of the antinode having the \((\mathrm{g})\) smallest, (h) second smallest, and (i) third smallest value of \(x\) ?

Use the wave equation to find the speed of a wave given by $$ y(x, t)=(3.00 \mathrm{~mm}) \sin \left[\left(3.00 \mathrm{~m}^{-1}\right) x-\left(8.00 \mathrm{~s}^{-1}\right) t\right] . $$

Two sinusoidal waves of the same frequency are to be sent in the same direction along a taut string. One wave has an amplitude of \(5.50 \mathrm{~mm}\), the other \(12.0 \mathrm{~mm}\). (a) What phase difference \(\phi_{1}\) between the two waves results in the smallest amplitude of the resultant wave? (b) What is that smallest amplitude? (c) What phase difference \(\phi_{2}\) results in the largest amplitude of the resultant wave? (d) What is that largest amplitude? (e) What is the resultant amplitude if the phase angle is \(\left(\phi_{1}-\phi_{2}\right) / 2 ?\)

A sinusoidal wave of angular frequency \(1200 \mathrm{rad} / \mathrm{s}\) and amplitude \(3.00 \mathrm{~mm}\) is sent along a cord with linear density \(4.00 \mathrm{~g} / \mathrm{m}\) and tension \(1200 \mathrm{~N}\). (a) What is the average rate at which energy is transported by the wave to the opposite end of the cord? (b) If, simultaneously, an identical wave travels along an adjacent, identical cord, what is the total average rate at which energy is transported to the opposite ends of the two cords by the waves? If, instead, those two waves are sent along the same cord simultaneously, what is the total average rate at which they transport energy when their phase difference is (c) 0 , (d) \(0.4 \pi \mathrm{rad}\), and (e) \(\pi \mathrm{rad}\) ?

The linear density of a string is \(1.9 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A transverse wave on the string is described by the equation $$ y=(0.021 \mathrm{~m}) \sin \left[\left(2.0 \mathrm{~m}^{-1}\right) x+\left(30 \mathrm{~s}^{-1}\right) t\right] \text {. } $$ What are (a) the wave speed and (b) the tension in the string?

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