Asked by Kilin
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]
[speed of sound in air=330m/s]
Answers
Answered by
Anonymous
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
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
Answered by
johnny
solution checks out but they made a typo in line 4 it should be 8.4 not 8.4^2
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.