Chapter 17: Problem 88
At a certain point, two waves produce pressure variations given by \(\Delta p_{1}=\Delta p_{m} \sin \omega t\) and \(\Delta p_{2}=\Delta p_{m} \sin (\omega t-\phi)\). At this point, what is the ratio \(\Delta p_{r} / \Delta p_{m}\), where \(\Delta p_{r}\) is the pressure amplitude of the resultant wave, if \(\phi\) is (a) 0, (b) \(\pi / 2\), (c) \(\pi / 3\), and (d) \(\pi / 4 ?\)
Short Answer
Step by step solution
Understanding the Problem
Finding the Resultant Pressure Amplitude
Calculation for \( \phi = 0 \)
Calculation for \( \phi = \frac{\pi}{2} \)
Calculation for \( \phi = \frac{\pi}{3} \)
Calculation for \( \phi = \frac{\pi}{4} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pressure Variations
- \( \Delta p_{1} = \Delta p_{m} \sin \omega t \)
- \( \Delta p_{2} = \Delta p_{m} \sin (\omega t - \phi) \)
Resultant Amplitude
- \[ \Delta p_{r} = 2 \Delta p_{m} \cos\left(\frac{\phi}{2}\right) \]
Phase Difference
- If \( \phi = 0 \), the waves are perfectly in phase, leading to a maximum constructive interference, resulting in the greatest possible amplitude.
- When \( \phi = \frac{\pi}{2} \), the waves are out of phase by a quarter cycle, causing partial interference that reduces the overall amplitude.
- Further, values like \( \phi = \frac{\pi}{3} \) and \( \phi = \frac{\pi}{4} \) create different levels of interference, modifying the amplitude accordingly.
Trigonometric Identities
- \[ \sin(x - \phi) = \sin x \cos \phi - \cos x \sin \phi \]
- \[ \Delta p_{r} = 2 \Delta p_{m} \cos\left(\frac{\phi}{2}\right) \]