Problem 44 - University Physics with Modern Physics

Question

Two speakers \(A\) and \(B\) are 3.50 m apart, and each one is emitting a frequency of 444 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. P35.44) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.

Step-by-Step Solution

Answer

The sound cancels at angles \( \pm 3.1^\circ, \pm 12.7^\circ, \ldots \) from the centerline.

Step 1: Calculate the Wavelength

First, we need to find the wavelength of the sound waves produced by the speakers. The formula for the wavelength \( \lambda \) is given by:\[ \lambda = \frac{v}{f} \]where \( v = 340 \text{ m/s} \) is the speed of sound and \( f = 444 \text{ Hz} \) is the frequency. Substitute the values to get:\[ \lambda = \frac{340}{444} \approx 0.766 \text{ m} \]

Step 2: Determine Phase Difference in Distance

Since speaker \( A \) is one-fourth of a period ahead, the initial phase difference, in terms of distance, is \( \frac{\lambda}{4} \). Given \( \lambda \approx 0.766 \text{ m} \), this equals:\[ \text{Phase difference} = \frac{0.766}{4} \approx 0.1915 \text{ m} \]

Step 3: Define the Condition for Destructive Interference

Destructive interference occurs when the path difference results in a phase difference of \((2m+1)\frac{\lambda}{2}\), where \(m\) is an integer (0, 1, 2,...). Thus, the condition for total path difference is:\[ \Delta x = (2m+1)\frac{\lambda}{2} - \frac{\lambda}{4} \approx 0.383m - 0.1915 \text{ m} \]

Step 4: Set Up and Solve the Equation for Path Difference

The effective path difference depends also on the separation between the speakers and angle \( \theta \) from the centerline. Use the equation:\[ d \sin\theta = (2m+1)\frac{\lambda}{2} - \frac{\lambda}{4} \]where \( d = 3.5 \text{ m} \). Substitute \( \lambda \approx 0.766 \text{ m} \) into the equation:\[ 3.5 \sin\theta = (2m+1)\times 0.383 \text{ m} - 0.1915 \text{ m} \]Solve for \( \theta \):\[ \theta = \arcsin\left(\frac{(2m+1)\times 0.383 - 0.1915}{3.5}\right) \]

Step 5: Calculate Angles for Different Values of m

To find all angles where destructive interference occurs, substitute integer values like \(m = 0, 1, 2,...\) until the \(\sin\theta\) exceeds 1 or becomes too negative. For each valid \(m\), calculate \(\theta\):- For \( m=0 \): \[ \theta = \arcsin\left(\frac{0.1915}{3.5}\right) \approx 3.1^\circ \]- For \( m=1 \): \[ \theta = \arcsin\left(\frac{0.766}{3.5}\right) \approx 12.7^\circ \]- Continue with higher \( m \) values until no solutions exist (\(|\sin\theta| \leq 1 \)).

Step 6: List the Angles on Both Sides of the Centerline

The angles found above represent the deviation from the centerline. For each positive \( \theta \), there is a corresponding negative angle. The cancellations occur symmetrically around the centerline at:\[ \theta = \pm 3.1^\circ, \pm 12.7^\circ, \ldots \]

Key Concepts

Sound Wave Interference

Sound wave interference is a fascinating phenomenon where two or more sound waves overlap and interact with each other. This interaction can lead to different effects such as constructive or destructive interference, depending on how the waves combine.
When it comes to destructive interference, as in the exercise provided, the waves interfere in such a way that they cancel each other out, resulting in a reduction or complete elimination of the sound. This occurs when the crest of one wave aligns with the trough of another, leading to a phase difference that causes cancellation.
Understanding interference is crucial when analyzing sound patterns, especially in fields related to acoustics, audio engineering, and physics.

Phase Difference

The phase difference between sound waves is a measure of how far the peaks or troughs of two waves are shifted relative to each other. It is crucial in determining whether waves will interfere constructively or destructively.
In the case of the original exercise, speaker A is ahead of speaker B by one-fourth of a period. This phase difference can be translated into a distance, especially when considering the path each sound wave travels. The initial phase difference in terms of distance was calculated as \(\frac{\lambda}{4}\), leading to a value of approximately 0.1915 meters.
This initial phase difference sets the stage for calculating conditions under which destructive interference will occur, aligning with peaks and troughs in such a way that they negate each other's presence.

Wavelength Calculation

A fundamental step in understanding sound wave behavior is calculating the wavelength. Wavelength \(\lambda\) is the distance over which the wave's shape repeats. It can be calculated using the formula \( \lambda = \frac{v}{f} \), where \(v\) is the speed of sound and \(f\) is the frequency of the wave.
For the exercise, the speed of sound was given as 340 m/s and the frequency as 444 Hz. Substituting these values, we get: \( \lambda = \frac{340}{444} \approx 0.766 \, \text{m} \).
Knowing the wavelength is vital for determining the phase and path differences that influence interference patterns. It helps in setting the mathematical groundwork for understanding how sound waves will interfere.

Path Difference

Path difference refers to the difference in the distances that two waves travel from their respective sources to a common point of interference. It's a key factor in determining whether the waves will interfere constructively or destructively.
In the original problem, the condition for destructive interference is influenced by the equation \( \Delta x = (2m+1)\frac{\lambda}{2} - \frac{\lambda}{4} \). This accounts for the initial phase difference and sets the criteria for complete cancellation of waves.
By manipulating this equation, one can solve for the angles where the interference pattern emerges, helping in identifying the positions where sound cancellation will occur. The path difference is thus essential in accurately modeling and predicting wave behavior in various applications.

Speaker Separation

The distance between sound sources, like speakers, plays a crucial role in how sound waves interact. In the context of the exercise, the speakers were separated by 3.5 meters, influencing how the sound waves from each source met and interacted.
This separation creates a specific intensity and pattern of interference, notably affecting the angles where destructive interference occurs. With a known wavelength, the separation determines the interference pattern along different angles, making it possible to calculate precisely where the sound waves will cancel out.
Understanding speaker separation in scenarios such as this one aids in designing sound systems efficiently and aligning speakers to optimize sound quality and coverage.