Problem 49 - University Physics with Modern Physics

Question

Two speakers \(A\) and \(B\) are 3.50 \(\mathrm{m}\) apart, and each one is emitting a frequency of 444 \(\mathrm{Hz}\) . However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\) . For points far from the speakers, find all the angles relative to the centerline (Fig. \(\mathrm{P} 35.49\) ) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) .

Step-by-Step Solution

Answer

Angles for destructive interference occur at \( \sin^{-1}(\pm 0.109 \times m) \) for odd \( m \).

Step 1: Calculate Wavelength

The wavelength \( \lambda \) of the sound can be found using the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound and \( f \) is the frequency. Substituting the values given, \( v = 340 \, \text{m/s} \) and \( f = 444 \, \text{Hz} \), we get: \[ \lambda = \frac{340}{444} \approx 0.766 \text{ m} \]

Step 2: Calculate Phase Difference

Speaker \( A \) is ahead of speaker \( B \) by one-fourth of a period, translating into a phase difference of \( \phi = \frac{1}{4}(2\pi) = \frac{\pi}{2} \).

Step 3: Conditions for Destructive Interference

For destructive interference (sound cancellation) between the two speakers, the condition \( \Delta \phi + k(2\pi) = n\pi \) must be met, where \( \Delta \phi \) is the phase difference and \( n \) is an odd integer. Substituting \( \Delta \phi = \frac{\pi}{2} \) into the condition, we get: \( \frac{\pi}{2} + k(2\pi) = (2m+1)\pi \), which simplifies to: \[ k = \frac{2m+1}{4} \] where \( m \) is an integer.

Step 4: Calculate Path Difference for Destructive Interference

Use the relationship for path difference in terms of angles: \( d \sin(\theta) = (k - \frac{1}{4})\lambda \). Substituting \( k = \frac{2m+1}{4} \) into the equation, the equation becomes: \[ d \sin(\theta) = \left( \frac{2m+1}{4} - \frac{1}{4} \right) \lambda = \frac{m \lambda}{2} \] with \( d = 3.50 \, \text{m} \).

Step 5: Solve for Angles Relative to the Centerline

Solving for \( \theta \): \[ \sin(\theta) = \frac{m \lambda}{2d} = \frac{m}{2} \times \frac{0.766}{3.50} \approx \frac{0.109m}{2} \] For it to be valid, \( |\sin(\theta)| \leq 1 \), which implies \( m = \pm1, \pm3, \pm5, \ldots \). Thus, calculate \( \theta \) for these \( m \) values:

Key Concepts

Phase Difference

When two wave sources, like speakers, produce sound at slightly different times, there is a difference in the phase of the waves they generate. This phase difference influences how sound waves interact with one another. In the original exercise, speaker A is a quarter of a period ahead of speaker B. This means there is a phase difference of \( \Delta \phi = \frac{\pi}{2} \). The phase difference is critical to determining how the waves will combine, as it determines whether they will enhance each other (constructive interference) or cancel each other out (destructive interference). Understanding the phase difference helps predict sound wave behavior, especially when the paths of the waves differ.

Destructive Interference

Destructive interference occurs when two sound waves combine to produce a wave of reduced or zero amplitude. This happens when the phase difference between the waves leads to a perfectly out-of-phase alignment, causing them to cancel each other out.In this exercise, we achieve destructive interference by setting the phase condition \( \Delta \phi + k(2\pi) = n\pi \), where \( n \) is an odd integer. This expression ensures that the sound waves generated by the two speakers meet in such a way that their peaks align with the troughs of the other, resulting in sound cancellation. This principle is particularly useful in applications requiring noise reduction, like in noise-canceling headphones.

Path Difference

The path difference is the physical difference in distance that two waves travel to reach the same point. This plays a vital role in determining interference patterns. For destructive interference, this path difference should satisfy certain conditions derived from the wave properties and their angles of travel.In the context of the given problem, the path difference involves spacing \( d \) between the speakers and angle \( \theta \) relative to the centerline. The equation \( d\sin(\theta) = \frac{m\lambda}{2} \) relates the path difference to the wavelength and ensures destructive interference for specific angles. This equation helps identify positions where the amplitudes of waves from both speakers will destructively interfere and cancel each other.

Wave Interference

Wave interference is the phenomenon where two or more waves superpose to form a resultant wave. This effect can be observed when waves overlap as they travel through a medium like air. There are two primary types of interference: constructive and destructive. This interplay between waves' paths and phases occurs naturally in many contexts, such as light and water waves, but is particularly significant in sound waves as studied in this problem. By understanding wave interference, one can manipulate sound to enhance or cancel specific frequencies, aiding in sound design and audio engineering.

Sound Waves

Sound waves are mechanical waves that require a medium to travel through, such as air, water, or solids. They are longitudinal waves, consisting of compressions and rarefactions moving through the medium.The frequency of the sound waves is key to sound perception—affecting pitch. In this exercise, the frequency of 444 Hz falls within the audible range for humans and dictates the wavelength through the relation \( \lambda = \frac{v}{f} \).Understanding sound waves involves considering not only their speed and frequency but how they interact with each other, as explored through concepts like phase and path difference. This comprehension is crucial for many technological applications, including acoustics, speech recognition, and broadcast systems.