Question

Two sources, \(A\) and \(B\), emit a sound of a certain wavelength. The sound emitted from both sources is detected at a point away from the sources. The sound from source \(\mathrm{A}\) is a distance \(d\) from the observation point, whereas the sound from source \(\mathrm{B}\) has to travel a distance of \(3 \lambda .\) What is the largest value of the wavelength, in terms of \(d\), for the maximum sound intensity to be detected at the observation point? If \(d=10.0 \mathrm{~m}\) and the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\), what is the frequency of the emitted sound?

Short answer

The largest value of the wavelength for the maximum sound intensity to be detected at the observation point is 2.5 meters, and the frequency of the emitted sound is 136 Hz.

Step-by-step solution

Define the given information

Based on the given information, we are given that source A is a distance \(d\) from the observation point, while source B is a distance of \(3\lambda\) from the observation point. Let's denote the path difference as \(\delta = d - 3\lambda\). We are also given the speed of sound, \(v = 340\,\mathrm{m/s}\).

Determine the condition for constructive interference

Constructive interference occurs when the waves from the two sources arrive at the observation point in phase. That is, the path difference \(\delta\) must be an integer multiple of the wavelength \(\lambda\). Mathematically, this can be written as: $$\delta = n\lambda$$ Where \(n\) is an integer.

Obtain an expression for the largest wavelength for maximum intensity

We need the largest value of \(\lambda\) to have constructive interference (maximum intensity) at the observation point, so we must take the smallest possible value of \(n\), which is \(n = 1\). We can write the equation as: $$d - 3\lambda = \lambda$$ To obtain an expression for the largest wavelength, we can rearrange the equation as follows: $$\lambda = \frac{d}{4}$$

Calculate the largest wavelength

Given that \(d = 10.0\,\mathrm{m}\), substitute this value into the expression for the largest wavelength: $$\lambda_{max} = \frac{d}{4} = \frac{10.0\,\mathrm{m}}{4} = 2.5\,\mathrm{m}$$

Calculate the frequency

We can use the speed of sound formula to determine the frequency of the emitted sound: $$f = \frac{v}{\lambda_{max}}$$ Substitute the given values, \(v = 340\,\mathrm{m/s}\) and \(\lambda_{max} = 2.5\,\mathrm{m}\): $$f = \frac{340\,\mathrm{m/s}}{2.5\,\mathrm{m}} = 136\,\mathrm{Hz}$$ The largest value of the wavelength for the maximum sound intensity to be detected at the observation point is \(2.5\,\mathrm{m}\), and the frequency of the emitted sound is \(136\,\mathrm{Hz}\).

Key concepts

Constructive Interference

Constructive interference happens when two or more waves meet and combine to form a wave of greater amplitude. This occurs when the waves are in phase, meaning their peaks and troughs match up perfectly. At the observation point, constructive interference results in the maximum sound intensity.

In mathematical terms, constructive interference is achieved when the path difference between two waves is an integer multiple of the wavelength. This can be expressed as \( \delta = n\lambda \), where \( \delta \) is the path difference, \( \lambda \) is the wavelength, and \( n \) is an integer.

• If the path difference is zero or a multiple of the wavelength, the waves reinforce each other.
• This results in a louder, more intense sound at the observation point.

In our example, the largest wavelength for maximum sound intensity can be found by considering the simplest case, \( n = 1 \), leading to the equation \( \lambda = \frac{d}{4} \). This maximum occurs when the sound waves travel paths that allow their peaks to emerge simultaneously.

Wavelength Calculation

The wavelength calculation is crucial to understanding wave behavior. It determines how far a wave travels in one complete cycle. In exercises involving wave interference, knowing the wavelength allows us to predict interference patterns and the resulting sound intensity.

The formula for wavelength in constructive interference scenarios is given by \( \lambda = \frac{d}{4} \), using the known path difference and required conditions for maximum intensity.

• This formula arises from setting the path difference to an integer multiple of the wavelength.
• Choosing the smallest integer, \( n = 1 \), simplifies the problem and gives the largest eligible wavelength.

This approach ensures the calculated wavelength leads to potent interference, maximizing sound intensity.

Sound Frequency

Sound frequency refers to the number of wave crests passing a specific point per second. It is measured in hertz (Hz) and is related to the speed of sound and wavelength through the equation \( f = \frac{v}{\lambda} \).

This formula indicates that as the wavelength changes, the frequency adjusts accordingly.

• The sound frequency will be higher with a shorter wavelength for a constant speed of sound.
• In our specific case, with a speed of sound \( v = 340 \, \mathrm{m/s} \) and a wavelength of \( 2.5 \, \mathrm{m} \), the frequency was calculated as \( 136 \, \mathrm{Hz} \).

Understanding sound frequency helps in tuning musical instruments and in designing acoustics, providing a fundamental bridge between theoretical physics and practical applications.

Wave Path Difference

Wave path difference is the difference in distance traveled by two waves from their respective sources to a common point. It is a critical factor in determining how these waves will interact.

A positive or negative wave path difference can lead to various interference patterns, such as constructive or destructive interference.

• A positive wave path difference indicates that one wave travels farther than the other.
• If this path difference is a whole multiple of the wavelength, constructive interference occurs.

In the context of the exercise, with source A's distance as \( d \) and source B's distance as \( 3\lambda \), the path difference is \( \delta = d - 3\lambda \). Matching this difference with an integer multiple of the wavelength ensures that waves reinforce each other, maximizing sound intensity at the observation point.