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Let's start with understanding how frequency calculation plays a role in sound interference. Frequency tells us how many oscillations per second a sound wave makes. It's crucial because it determines the pitch of the sound. In this scenario, you're dealing with sound waves coming from two loudspeakers. To find the frequency, you need to use the formula:\[ v = f \lambda \]where \( v \) is the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength. Given that the speed of sound is approximately \(343\ \text{m/s}\), and the wavelength was deduced to be \(0.5\ \text{m}\), you can calculate the frequency as follows:\[ f = \frac{v}{\lambda} = \frac{343\ \text{m/s}}{0.5\ \text{m}} = 686\ \text{Hz} \]This result, \(686\ \text{Hz}\), is the frequency where you initially observe a minimum in sound intensity.
It's essential to remember this frequency, as it will help in further steps when determining conditions for maximum sound intensity.

Sound intensity is essential in understanding sound wave interference. It measures how much sound energy passes through a certain area. Imagine standing equidistant between two loudspeakers. Here, sound waves interfere with each other, creating areas of high and low intensity.
When sound waves constructively interfere—they enhance each other's effect, creating a "maximum" of sound intensity. Conversely, destructive interference leads to a "minimum" of sound intensity, where waves cancel each other out.

• At a maximum, the waves are in phase.
• At a minimum, they are out of phase.

In this exercise, you first noted a minimum intensity at a displacement of \(0.25\ \text{m}\) from the midway point—an indicator of destructive interference. By changing the sound frequency, you influence where these maxima and minima occur.
This exercise demonstrates the positions along the line between the speakers where constructive or destructive interference will cause peaks or nulls in sound intensity.

Understanding how wavelength is determined is key in sound wave problems. The wavelength \( \lambda \) is the distance between successive peaks of a wave. It's related to the wave's speed and frequency by the equation \( v = f \lambda \). For sound waves in air, the speed \( v \) is approximately \(343\ \text{m/s}\). In this exercise, when you first notice a minimum of sound intensity at \(0.25\ \text{m}\) from the center, it gives you a clue about the wavelength of the sound.- At a minimum of intensity, the path difference is half of the wavelength.
- Thus, the wavelength \( \lambda \) can be calculated as: \[ \lambda = 2 \times \text{path difference} = 2 \times 0.25\ \text{m} = 0.5\ \text{m} \]Why is calculating wavelength important? Knowing \( \lambda \) not only allows us to find frequency but also helps determine the spatial distribution of interference patterns.
When the frequency changes, affecting \( \lambda \), the pattern of sound intensity shifts, affecting where maxima and minima will occur. With a new maximum, you discover the location's conditions for optimal sound experience.