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 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

Answer: The frequency of the emitted sound is approximately 102 Hz.

Step-by-step solution

Analyzing Constructive Interference Condition

For maximum sound intensity to be observed, the path difference between the two sound sources and the observation point should be an integral multiple of the wavelength. Mathematically this can be expressed as: \((n)\lambda = d - 3\lambda\), where n is an integer. This equation represents the condition for constructive interference, which leads to maximum sound intensity.

Solve for the Wavelength

Now, we'll rearrange the equation from Step 1 to find the largest value of the wavelength in terms of d: \(\lambda (n + 3) = d\) \(\lambda = \frac{d}{n + 3}\) Since we need to find the largest value of the wavelength, and n is an integer, we should choose the smallest possible value for n, which is 0: \(\lambda = \frac{d}{0 + 3} = \frac{d}{3}\)

Calculate the Frequency of the Emitted Sound

Now that we have found the largest value of the wavelength in terms of \(d\), we can plug in the given speed of sound (\(v = 340m/s\)) and distance (\(d = 10.0m\)) to determine the frequency of the emitted sound: \(\lambda = \frac{d}{3} = \frac{10}{3} = 3.33\mathrm{m}\) Using the wave equation, \(v = f\lambda\), we can isolate frequency \(f\): \(f = \frac{v}{\lambda} = \frac{340}{3.33} = 102\mathrm{Hz}\) So, the frequency of the emitted sound is approximately 102 Hz.

Key concepts

Sound Intensity

Sound intensity can be thought of as the power carried by sound waves per unit area, which is a measure of how much energy passes through a certain area in a certain amount of time. When two sound waves meet, they can interfere with each other, and if they are in phase, they can augment each other's effects - a phenomenon known as constructive interference, resulting in increased sound intensity.

Imagine two speakers facing each other and playing the same note. At some points between them, the sounds waves will meet in sync, and you'll hear a louder sound; that’s because the waves are reinforcing each other, adding their energies together to create a higher intensity. This is precisely what occurs at the observation point in the exercise when the path difference is an integral multiple of the wavelength.

Wavelength

Wavelength is the distance between consecutive corresponding points of the same phase on a wave, such as from crest to crest or trough to trough. It is typically denoted by the Greek letter lambda \( \lambda \). In the context of the exercise, the wavelength is a key component in determining how sound waves from different sources will interfere with each other. The larger the wavelength in relation to the distance \(d\), the lower the frequency of the sound. For two points to be in constructive interference, the path length difference must equal an integer multiple of the wavelength.

When solving for the wavelength that results in the maximum sound intensity, we find that a smaller integer \(n\) in the formula \( \lambda(n + 3) = d \) will yield a larger wavelength. Thus, the selection of \(n = 0\) gives us the maximum wavelength possible for constructive interference to occur.

Frequency of Sound

Frequency, often denoted as \(f\), is the number of oscillations or cycles that a wave undergoes per unit of time, and is commonly expressed in Hertz (Hz). The frequency of sound determines its pitch; a higher frequency implies a higher pitch, while a lower frequency results in a lower pitch.

In the exercise, once the wavelength \( \lambda \) is determined, it can be used along with the wave equation to find the frequency of the sound emitted from sources A and B. The wave equation links the speed of sound (in this case, \(340 \text{m/s}\)), the frequency, and the wavelength together. By rearranging this equation \(v = f\lambda\), we can solve for the frequency, revealing a direct inverse relationship between frequency and wavelength. As the wavelength increases, the frequency decreases, which means for the maximum wavelength determined, we will have a corresponding minimum frequency that the sources can emit to still observe constructive interference at the detection point.

Wave Equation

The wave equation represents a fundamental relationship between the speed of a wave \(v\), its frequency \(f\), and its wavelength \(\lambda\): \[ v = f\lambda \
It is the backbone of understanding how waves move and how their various properties are interconnected. In the context of the exercise, this equation forms the bridge between the determined wavelength and the frequency we are looking to find.

By rearranging to solve for frequency, \(f = \frac{v}{\lambda}\), we apply the known values for the speed of sound in air and the calculated wavelength to find the emitting sources' frequency. It's crucial to recognize that the speed of sound can be affected by factors such as temperature and humidity, but in a controlled environment, or when average conditions are assumed, the wave equation provides an essential tool for solving many acoustical physics problems.