Problem 3 - University Physics with Modern Physics

Question

Two speakers that are 15.0 m apart produce in-phase sound waves of frequency 250.0 Hz in a room where the speed of sound is 340.0 m/s. A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

Step-by-Step Solution

Answer

(a) Constructive; (b) 0.34 m; (c) 0.68 m.

Step 1: Calculate the Wavelength

First, we find the wavelength of the sound using the formula \( \lambda = \frac{v}{f} \) where \( \lambda \) is the wavelength, \( v = 340.0 \, \text{m/s} \) is the speed of sound, and \( f = 250.0 \, \text{Hz} \) is the frequency. Thus, \( \lambda = \frac{340.0}{250.0} = 1.36 \, \text{m} \).

Step 2: Determine Initial Interference Type

At the midpoint, the woman is equidistant from both speakers. Since the speakers are in-phase, the path difference is zero at the midpoint. When the path difference is zero, constructive interference occurs. Therefore, she hears constructive interference.

Step 3: Calculate Path Difference for Destructive Interference

Destructive interference occurs when the path difference is \( \frac{\lambda}{2}, \frac{3\lambda}{2}, \ldots \). For the first instance of destructive interference, the path difference should be \( \frac{\lambda}{2} = \frac{1.36}{2} = 0.68 \, \text{m} \).

Step 4: Find Distance for Minimum Intensity

Considering she walks toward one speaker, to achieve a path difference of \( 0.68 \, \text{m} \), she must walk \( 0.34 \, \text{m} \) from the midpoint toward one speaker. Since the woman starts at the midpoint where both distances are equal, walking \( 0.34 \, \text{m} \) changes the distance between the speakers by \( 0.68 \, \text{m} \).

Step 5: Calculate Path Difference for Constructive Interference

Constructive interference occurs when the path difference is a multiple of \( \lambda \), hence either zero or \( \lambda \), etc. Since the midpoint is already a location of constructive interference, to find the next point, the path difference should be \( \lambda = 1.36 \, \text{m} \).

Step 6: Find Distance for Maximum Intensity

The woman must again change the distance between her and one speaker from the midpoint such that the additional distance is a wavelength \( 1.36 \, \text{m} \). To achieve this separation, she needs to walk \( 0.68 \, \text{m} \), since the path difference doubles her walk from the center.

Key Concepts

Constructive Interference

Constructive interference is a fascinating phenomenon that occurs when two or more sound waves meet and combine to produce a wave with greater amplitude. This situation happens when the waves are in phase, meaning they crest and trough together. When the woman stands at the midpoint between two in-phase speakers, she experiences constructive interference.
At this point, the sound waves from both speakers travel the same distance to reach her. The path difference, or the difference in distance that each wave travels, is zero. This zero path difference ensures that the waves reinforce each other, resulting in a louder sound.

• The waves must arrive in phase.
• Their amplitudes add constructively.
• Resulting in increased sound intensity.

The midpoint, therefore, is where the woman will first experience maximum sound intensity due to constructive interference.

Destructive Interference

Destructive interference is the opposite of constructive interference. It occurs when two sound waves meet out of phase, meaning they cancel out each other’s amplitude. This results in a wave with reduced amplitude, or in complete cases, no sound at all.
For destructive interference, the path difference must be a half-multiple of the wavelength, such as \( \frac{\lambda}{2}, \frac{3\lambda}{2} \). In this scenario, when the woman walks \( 0.34 \text{ m} \) towards one speaker, the path difference becomes \( 0.68 \text{ m} \), half of the calculated wavelength \( 1.36 \text{ m} \). At this point, she hears the sound intensity dip to a minimum.

• Occurs when waves are out of phase.
• Wave amplitudes subtract, reducing intensity.
• Path difference results in minimal sound.

Path Difference

Path difference is a critical concept in understanding sound wave interference. It refers to the difference in distance traveled by different waves from their source to the point of interference. The path difference determines whether interference will be constructive or destructive.
For constructive interference, the path difference is a multiple of the wavelength \( \lambda \). As seen in the exercise, the woman experiences constructive interference at the midpoint where the path difference is zero. For the next instance of constructive interference along her path, the path difference should be \( 1.36 \text{ m } \), the whole wavelength.
In destructive interference, the path difference is a half-multiple of the wavelength, \( \frac{\lambda}{2} \). This is why walking \( 0.34 \text{ m} \) from the midpoint leads to the first occurrence of minimal sound, with a path difference of \( 0.68 \text{ m} \).

• Path difference is key to interference type.
• Zero/multiple \( \lambda \): Constructive.
• Half-multiple \( \lambda \): Destructive.

Sound Wave Frequency

The frequency of a sound wave is an essential factor in determining the nature of interference. Frequency, denoted by \( f \), is the number of wave cycles per second and is measured in Hertz (Hz). In this exercise, each speaker produces sound waves at a frequency of 250.0 Hz.
This frequency, combined with the speed of sound \( 340.0 \text{ m/s} \), allows us to calculate the wavelength \( \lambda \) of the sound, which is \( 1.36 \text{ m} \).

• High frequency often means shorter wavelength.
• Wavelength calculation: \( \lambda = \frac{v}{f} \).
• Wavelength affects interference conditions.

Sound frequency and stage of interference go hand in hand as both inform how sound reacts in a space, ultimately influencing where the woman experiences peaks and troughs of sound intensity.