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Chapter 11: Problem 35

A traveling sine wave is the result of the superposition of two other sine waves with equal amplitudes, wavelengths, and frequencies. The two component waves each have amplitude \(5.00 \mathrm{cm} .\) If the superposition wave has amplitude \(6.69 \mathrm{cm},\) what is the phase difference \(\phi\) between the component waves? [Hint: Let \(y_{1}=A \sin (\omega t+k x)\) and $y_{2}=A \sin (\omega t+k x-\phi) .$ Make use of the trigonometric identity (Appendix A.7) for \(\sin \alpha+\sin \beta\) when finding \(y=y_{1}+y_{2}\) and identify the new amplitude in terms of the original amplitude. \(]\)

Short Answer

Expert verified
Answer: The phase difference between the two component waves is approximately 96.4 degrees.

Step by step solution

01

Write the given equations

The given equations for the component waves are: \(y_1 = A \sin (\omega t + kx)\) and \(y_2 = A \sin (\omega t + kx - \phi)\) where A is the amplitude, k is the wave number, \(\omega\) is the angular frequency, and \(\phi\) is the phase difference between the two waves.
02

Add the component waves to find the resultant wave

To find the resultant wave, we need to add the component waves: \(y = y_1 + y_2\) \(y = A\sin(\omega t + kx) + A\sin(\omega t + kx - \phi)\)
03

Use the trigonometric identity for adding sine functions

We use the trigonometric identity for adding sine functions: \(\sin{\alpha}+\sin{\beta}=2\sin{\frac{\alpha + \beta}{2}}\cos{\frac{\alpha - \beta}{2}}\) Applying this identity to the equation for the resultant wave: \(y = 2A\sin{\frac{(\omega t + kx) + (\omega t + kx - \phi)}{2}}\cos{\frac{(\omega t + kx) - (\omega t + kx - \phi)}{2}}\) Simplify the equation: \(y = 2A\sin({\omega t + kx - \frac{\phi}{2}})\cos{-\frac{\phi}{2}}\)
04

Write amplitude relationship and solve for φ

We are given that the amplitude of the resultant wave is 6.69 cm. Therefore, the amplitude of the superposition wave can be written as: \(6.69 \mathrm{cm} = 2(5.00 \mathrm{cm})\cos{-\frac{\phi}{2}}\) Solve for \(\cos{-\frac{\phi}{2}}\): \(\cos{-\frac{\phi}{2}} = \frac{6.69 \mathrm{cm}}{2(5.00 \mathrm{cm})} = 0.669\) Now solve for the phase difference φ: \(\phi = -2 \cos^{-1}(0.669)\) \(\phi \approx 2(48.2^\circ)\) \(\phi \approx 96.4^\circ\) The phase difference between the two component waves is approximately 96.4 degrees.

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Most popular questions from this chapter

The speed of sound in air at room temperature is \(340 \mathrm{m} / \mathrm{s}\) (a) What is the frequency of a sound wave in air with wavelength $1.0 \mathrm{m} ?$ (b) What is the frequency of a radio wave with the same wavelength? (Radio waves are electromagnetic waves that travel at $3.0 \times 10^{8} \mathrm{m} / \mathrm{s}$ in air or in vacuum.)
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