Problem 37 - University Physics with Modern Physics

Question

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 860 Hz. Point \(P\) is 12.0 m from \(A\) and 13.4 \(\mathrm{m}\) from \(B\) . Is the interference at \(P\) constructive or destructive? Give the reasoning behind your answer.

Step-by-Step Solution

Answer

The interference at P is destructive because the path difference is an odd half-wavelength multiple.

Step 1: Identify Wavelength

The speed of sound in air is approximately 343 m/s. To find the wavelength of the sound waves, use the formula \( \lambda = \frac{v}{f} \), where \( v \) is the speed of sound and \( f \) is the frequency. \( \lambda = \frac{343}{860} \approx 0.399 \mathrm{\ m} \).

Step 2: Calculate Path Difference

The path difference between waves from the two speakers is the difference in the distances from point \(P\) to each speaker. Calculate this as: \( \text{Path Difference} = |13.4 - 12.0| = 1.4 \mathrm{\ m} \).

Step 3: Determine Constructive or Destructive Interference

For constructive interference, the path difference should be an integer multiple of the wavelength (\( m\lambda \)), and for destructive interference, it should be an odd multiple of half-wavelengths (\((m + \frac{1}{2})\lambda\)). Divide the path difference by the wavelength: \( \frac{1.4}{0.399} \approx 3.51\). This is not an integer, but close to 3.5, suggesting it is an odd multiple of half-wavelengths and thus destructive interference.

Key Concepts

Sound Waves

Sound waves are a type of mechanical wave that travels through a medium, typically air. Unlike electromagnetic waves, they require a medium to propagate. Sound waves are generated by vibrating objects which create pressure variations in the medium. As these pressure variations travel through the medium, they are perceived as sound by our ears.

Key characteristics of sound waves include:

• Wavelength: This is the distance between two consecutive points that are in phase, like two adjacent crests or troughs.
• Frequency: This refers to the number of wave cycles that pass a point per unit of time, measured in hertz (Hz).
• Speed: In air, sound travels at approximately 343 meters per second, although this can vary with temperature and humidity.

Sound waves can be transformed by various conditions, such as reflecting off surfaces or bending around obstacles, which is part of the phenomenon known as wave interference.

Constructive Interference

Constructive interference occurs when two or more waves meet in such a way that their crests overlap, resulting in a new wave of greater amplitude. This happens when the waves are in phase, meaning their peaks align perfectly.

Here’s how constructive interference works:

• The path difference between interacting waves results in a whole number of wavelengths, denoted as \(m\lambda\), where \(m\) is an integer.
• In such cases, the amplitude of the resulting wave is the sum of the individual wave amplitudes, leading to louder or more intense sounds.
• Constructive interference is commonly seen in scenarios where sound, light, or water waves add together.

This principle is essential in various applications, such as sound engineering, where phase alignment is used to amplify audio signals.

Destructive Interference

Destructive interference occurs when two or more waves meet and their crests and troughs are out of phase, effectively canceling each other out. This results in a wave of reduced or zero amplitude.

Here’s how you can identify destructive interference:

• The path difference between waves is an odd multiple of a half-wavelength, expressed as \((m + \frac{1}{2})\lambda\), where \(m\) is an integer.
• When waves are exactly half a wavelength apart, their crests meet troughs, causing cancellation and a pronounced reduction in sound intensity.
• This phenomenon is critical in various fields, such as acoustics, where it helps in noise cancellation technologies.

In the exercise example, the interference at point \(P\) is destructive because the path difference calculated resulted in an odd multiple of half-wavelengths, leading to cancellation of sound at that specific location.