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\(\bullet\) \(\bullet\) Two speakers that are 15.0 \(\mathrm{m}\) apart produce in- phase sound waves of frequency 250.0 \(\mathrm{Hz}\) in a room where the speed of sound is 340.0 \(\mathrm{m} / \mathrm{s} .\) 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

01

Calculate Wavelength

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

02

Determine Path Difference

The woman starts at the midpoint, equidistant from both speakers, so initially, the path difference is \(0\), resulting in in-phase (constructive) interference since both speakers emit sound waves in-phase.

03

Minimum Intensity Path Difference

For the sound to be at a minimum intensity (destructive interference), the path difference must be an odd multiple of \( \frac{\lambda}{2} \). The smallest non-zero path difference for this is \( \frac{\lambda}{2} = \frac{1.36}{2} = 0.68 \, \text{m} \).

04

Calculate Distance for Minimum Intensity

Since the path difference is due to the woman walking towards one speaker, let her walk a distance \( x \). The new path difference becomes \( x - (-x) = 2x \). Set \( 2x = 0.68 \) to find \( x \).\[ x = \frac{0.68}{2} = 0.34 \, \text{m} \]

05

Maximum Intensity Path Difference

For maximum intensity (constructive interference), the path difference must be a whole multiple of \( \lambda \). The smallest non-zero path difference for this situation is \( \lambda = 1.36 \, \text{m} \).

06

Calculate Distance for Maximum Intensity

Using the same reasoning as before, the path difference \( 2x \) should be \( 1.36 \, \text{m} \):\[ 2x = 1.36 \, \Rightarrow \, x = \frac{1.36}{2} = 0.68 \, \text{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sound Waves

Sound waves are a type of mechanical wave that travels through a medium, such as air, by means of vibrations. These vibrations cause the air particles to oscillate back and forth, transmitting energy from one location to another.

• Characteristics: Sound waves have several key characteristics, including frequency, wavelength, amplitude, and speed.
• Frequency: This is the number of oscillations or vibrations per second and is measured in Hertz (Hz). In the given problem, the frequency is 250 Hz.
• Speed: The speed of sound in air is approximately 340 m/s, although it can vary slightly depending on temperature and atmospheric conditions.
• Wavelength (λ): This is the distance between successive crests or troughs in a wave, calculated with the formula: \[ \lambda = \frac{v}{f} \]where \( v \) is wave speed and \( f \) is frequency.

Understanding these properties helps us predict how sound will behave, especially in interference scenarios.

Constructive Interference

Constructive interference occurs when two or more waves meet and combine to form a wave with a higher amplitude. For sound waves, this means the sound becomes louder. This typically takes place when the waves are in-phase, meaning the peaks (or compressions in a sound wave) align perfectly.

• In-phase Waves: For in-phase waves, the path difference between them should be a multiple of the wavelength. In the problem, starting at the midpoint, the waves are in-phase, leading to constructive interference.
• Maximum Intensity: The woman will hear the maximum sound intensity when the path difference equals an integer multiple of the wavelength (1, 2, 3 ... λ). Using the equation:\[ 2x = n \times \lambda \]she will first hear constructive interference after walking 0.68 m, the smallest distance satisfying this condition.

Destructive Interference

Destructive interference happens when two waves meet and cancel each other out, resulting in a wave with a lower amplitude. For sound waves, this means the sound gets quieter or even silent if the cancellation is complete. This occurs when waves are exactly out of phase.

• Out-of-phase Waves: For waves to be out-of-phase, their path difference must equal an odd multiple of half the wavelength. In this case, \[ \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, ... \]
• Minimum Intensity: In the exercise, the smallest distance for the woman to hear destructive interference is when the path difference equals 0.68 m. Using the formula:\[ 2x = \left( 2n + 1 \right) \times \frac{\lambda}{2} \]This means she needs to walk 0.34 m towards any speaker to first encounter a minimum in sound intensity.

Wavelength Calculation

To find the wavelength of a sound wave, you can use the formula: \[ \lambda = \frac{v}{f} \]where \( v \) is the speed of sound and \( f \) is the frequency of the sound wave. This calculation is crucial for predicting interference patterns.

• Example Calculation: In the given problem, the speed of sound \( v = 340 \) m/s and frequency \( f = 250 \) Hz. By substituting these values into the formula, we get:\[ \lambda = \frac{340}{250} = 1.36 \, \text{m} \]
• Importance of Wavelength: Knowing the wavelength allows you to determine points of constructive and destructive interference based on the path differences encountered. It helps predict where the peaks and troughs of sound waves will align or cancel out.

Understanding how to calculate and use wavelength is essential for solving problems related to wave interference.