Two fire alarms are placed in a building, one 2.3 m north of the other. A person sitting at a desk which is 67.2 m east and 8.4 m north of the point mid-way between the alarms does not hear them when they sound. What is the lowest frequency that the alarms can be sounding at assuming that the alarms are both perfectly in phase?

[speed of sound in air=330m/s]

solution checks out but they made a typo in line 4 it should be 8.4 not 8.4^2

Well, it seems like those fire alarms just don't want to be heard! Maybe they should consider taking some voice lessons? Anyway, let's do some calculations to find out the lowest frequency they can be sounding at.

First, let's find the distance between the person at the desk and each alarm. The distance from the person to the mid-way point between the alarms can be found using the Pythagorean theorem:
sqrt((67.2/2)^2 + 8.4^2)

Now, we subtract the distance of the alarm that is 2.3 m north of the other from this distance to get the total distance traveled by the sound waves:
sqrt((67.2/2)^2 + 8.4^2) - 2.3

Finally, we divide this total distance by the speed of sound in air to find the time it takes for the sound to reach the person:
(sqrt((67.2/2)^2 + 8.4^2) - 2.3) / 330

Now here comes the funny part... Just kidding, there's no punchline! We're almost there.

Since the alarms are perfectly in phase, the time it takes for the sound waves from both alarms to reach the person should be the same. This means that the time it takes for the sound waves to reach the person from each alarm should be equal.

So, we can set up an equation:
Time taken for sound wave from first alarm = Time taken for sound wave from second alarm

By substituting the expression we found for time in terms of distance and speed of sound, we get:

(sqrt((67.2/2)^2 + 8.4^2) - 2.3) / 330 = (sqrt((67.2/2)^2 + 8.4^2) + 2.3) / 330

Now, all we have to do is solve for the lowest frequency. But hey, just out of curiosity, have you heard about the fire alarm that went to school? It finally learned how to make some sound waves! Okay, I'll keep joking to a minimum now.

By solving the equation above for the lowest frequency, we can find the answer you're looking for. I hope this explanation wasn't alarmingly confusing!

To solve this problem, we need to calculate the distance between the person and each alarm, and then determine the time it takes for the sound to travel from each alarm to the person.

Let's go step by step:

1. Find the distance between the person and each alarm:
- The person is 67.2 m east and 8.4 m north of the midpoint between the alarms.
- Since the alarms are 2.3 m apart, the distance from the person to the alarm north of them is 8.4 m + 1.15 m (half of 2.3 m) = 9.55 m.
- The distance from the person to the alarm south of them is also 9.55 m.

2. Calculate the time for the sound to travel from each alarm to the person:
- Recall that the speed of sound in air is 330 m/s.
- The time it takes for the sound to travel is given by the formula: time = distance / speed.
- For the alarm north of the person: time = 9.55 m / 330 m/s = 0.029 secs.
- For the alarm south of the person: time = 9.55 m / 330 m/s = 0.029 secs.

3. Determine the frequency of the alarms:
- To be perfectly in phase, the difference in the time it takes for the sound to reach the person from each alarm must be a whole number of cycles.
- Since the alarms are in phase, the time difference between them is 0 seconds.
- Therefore, the time for the sound to travel from each alarm must be the same.
- The frequency of sound is given by the formula: frequency = 1 / time.
- The lowest frequency occurs when the time is the maximum, which is 0.029 secs.
- Therefore, the lowest frequency is 1 / 0.029 secs = 34.48 Hz.

So, the lowest frequency that the alarms can be sounding at is approximately 34.48 Hz.

half a wavelength path difference

8.4 - 1.15 = 7.25
d1^2 = 67.2^2 + 7.25^2
d1 = 67.589958

8.4^2 + 1.15 = 9.55
d2^2 = 67.2^2 + 9.55^2
d2 = 67.875198

d2 - d1 = .2852 meters
which is half a wavelength
so wavelength = .5705

.5705 = 330 m/s * 1/f
so
f = 578 Hz