An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Outside the door, on a line perpendicular to the opening in the door, a jet engine makes a \(600-\mathrm{Hz}\) sound. At what angle with the door will the technician observe the first minimum in sound intensity if the vertical opening is \(0.800 \mathrm{m}\) wide and the speed of sound is \(340 \mathrm{m} / \mathrm{s} ?\)

Understanding wave interference is vital to explaining many phenomena in physics, including the scenario described in our exercise. Interference occurs when two or more waves overlap and combine to form a new wave pattern. This principle is utilized in the analysis of sound, light, and water waves, among others. In the context of sound waves, which is our focus, interference can result in either constructive interference, where wave crests align to create a louder sound, or destructive interference, which can lead to quieter sounds or even silence.

In the exercise, we're dealing with destructive interference, which specifically happens in our single-slit scenario. Here, the sound waves spread out after passing through the hangar door (the slit) and begin to overlap in certain areas, causing some regions where the waves cancel each other out. This cancellation is what creates the minima in sound intensity that the technician experiences.

The formula \(sin(\theta) = \frac{m\lambda}{a}\) elegantly captures this relationship. In this formula, 'm' indicates the order of the minimum, which represents how many 'dark fringes' or points of destructive interference occur. Understanding this concept helps explain why the technician would observe a drop in sound intensity at specific angles relative to the door.

The wavelength of a sound wave is the physical distance between two consecutive crests or troughs of a wave. It is a critical parameter in acoustics, as it determines the pitch or tone we perceive. Longer wavelengths correspond to lower-pitched sounds, while shorter wavelengths yield higher-pitched sounds.

The wavelength \(\lambda\) of a wave can be found using the speed of the wave (v) and its frequency (f), as shown in the formula \(\lambda = \frac{v}{f}\). In the context of the exercise, the technician hears a 600 Hz sound, and given the speed of sound is 340 meters per second, the wavelength can be calculated. This information is crucial because it allows us to use the single-slit interference formula to determine at what angles minima will occur.

Understanding this concept not only supports the technical aspects of acoustics but also deepens appreciation for the everyday experience of sound. For example, knowing how to calculate wavelength allows one to design spaces with optimal acoustics for specific sounds or frequencies.

Diffraction is a phenomenon that occurs when a wave encounters an obstacle or passes through a narrow opening, causing the waves to spread out and alter direction. For sound waves, diffraction is the reason we can hear noises around corners or through openings even when the source is not in direct line of sight.

In our exercise, the hangar door acts as a single slit causing diffraction of the sound waves originating from the jet engine. The size of the slit relative to the wavelength of the sound influences how much the sound spreads. Generally speaking, when the slit is comparable in size to the wavelength of the sound, significant diffraction occurs. This is why the hangar door’s 0.800m width being comparable to the calculated wavelength of the sound (0.5667m) causes notable diffraction.

Understanding diffraction is integral to many applications, from architectural acoustics (designing buildings with desired sound properties), to the development of sound-based medical imaging techniques. It also underpins fundamental scientific principles used in a variety of technological innovations, such as the design of loudspeakers and microphone arrays.