As sound radiates from a point source, it does so in concentric circles. Therefore, we can calculate the intensity of sound using the following formula:

We simply need to calculate : your distance from the space shuttle.

The shuttle is 500 meters from the ground, and you are 300 meters from its launch point. We can calculate the distance from you to the current position of the shuttle by using the Pythagorean Theorem:

There is no need to take the square root, since it's already in the form we need to do the calculation. Plugging in this value to the original equation, we get:

A harmonic is a standing wave that has certain points (nodes) which do not go through any displacement, and certain points (antinodes) that are moving through maximum displacement during the wave's resonation.

Wavelength is related to harmonic on a string fixed at both ends through the formula , where  is the length of the string and  is the number harmonic. By plugging in the given length and wavelength, we can solve for .

The string must be vibrating at the fourth harmonic.

First, notice that the paragraph above tells us that the wave can be modeled as a pipe with one end closed. This is in contrast to the other possibility, where the wave is modeled as a pipe with two ends open. It is critical to recognize this difference, as the definition of sequential harmonics and the formula used to calculate them changes depending on whether both ends are open or not. In the situation where one end is closed, the harmonics are odd numbers, meaning that the first three harmonics are 1^st, 3^rd, and 5^th. In the situation where both ends are open or both ends are closed, the harmonics are sequential, meaning that the first three harmonics are 1^st, 2^nd, and 3^rd.

This question asks us to incorporate information we learned in the pre-question text and new information in the question. First, we are told that we are looking to calculate a distance where maximal sound is heard, in other words, where the amplitude is at its maximum. This only occurs at an anti-node. 

Additionally, we know that the speaker can be modeled as a one-end closed pipe, meaning that the wavelengths of the harmonics are calculated as .

In a one-end closed pipe model, each harmonic ends with maximal amplitude, so we can find L (the length of the pipe, i.e. where the person standing would hear maximal sound), as we know the fundamental harmonic has n = 1.

In an open-closed pipe, we get only odd harmonics, since there must always be a displacement node at the closed end and a displacement antinode at the open end. So, the first overtone is the same as the third harmonic, which has . Similarly, the second overtone is the fifth harmonic, which has . Using the given value of  in the first overtone to be 300Hz, we find  and .

Two waves will emit a beat with a fequency equal to the difference in frequency of the two waves. In this case the beat frequency is:

    (beats per second)

Convert to beats per minute:

First, let’s look at what this question is asking us to consider. The question wants us to determine the relationship between waves produced from our sine phase base speaker and a cosine phase speaker accidently placed directly across from it. Two important facts come to mind when thinking about this scenario. First, the waves physically overlap in space because the speakers are pointed directly at each other. Second, the sine and cosine waves are exactly 180º out of phase. Remember that waves completely out of phase create destructive interference, meaning that the resultant wave will have no amplitude. The volume directly between the two speakers is zero because volume is directly correlated to wave amplitude.

Two waves will cancel to zero amplitude when the relative shift between them is half a period. This corresponds to half of , which gives the correct answer of .

Destructive inference occurs when the peak of one wavelength encounters the trough of another wavelength. This causes the waves to cancel each other out, essentially producing a much smaller audible wavelength. In contrast, constructive interference will result in the summation of the wave amplitudes, and an increase in the volume and intensity of the sound.

The bright lines from a diffraction grating can be located with the equation , where  is wavelength, d is the distance between slits in the diffraction grating, and  is the angle of separation from the center of the interference pattern on the screen.   is proportional to , so if  decreases, the angle of separation between lines also decreases.