I'm unsure how to even start this problem or how I'm suppose to find out the distance at which destructive interference would occur...
Two loudspeakers are 1.8 m apart. A person stands 3.0 m from one speaker and 3.5 m from the other. (a) What is the lowest frequency at which destructive interference will occur at this point? (b) Calculate two other frequencies that also result in destructive interference at this point (give the next two highest). Let T = 20 degrees C.
so velocity of sound in air at 20 degrees C is 343 m/s
I don't even know were to go from here
see question 48 as an example
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PST4_2/T4_2.htm
everything made sense I just don't see why .5 m must be half a wave length
Also is the 343 hz the first harmonic or what is it because I know once you find the fundemental then the first harmonic is just simply twice the fundemental and the third harmonic is just simple three times the fundemental frequency...
I'm doing 343*3 to the get the next highest one which dosen't make sense coulding I find the second harmonic if 343 is the fundemental by multiplying 343 by 2 which would give me the second harmonic
wouldn't the second harmoinc be the next highest after the fundemental...
THANKS!
please if someone else knows...
YOu set the half wavelength to what ever it is .
Second harmonic is to most, the 2xfundamental.
However, I have to tell you, there are some that use first harmonic to mean 2x. This is from the musical usage of the term first overtone, which is 2xfundemantal.
The usage is changing finally, so most agree that they mean second harmonic is 2xfundamental. But be cautious on it, some folks mean different things.
idk !!!! im having trouble on the same question !!!! GOT NOO IDEA !!
To solve this problem, we need to understand the concept of interference and the conditions required for destructive interference to occur. Interference is the phenomenon that occurs when two or more waves combine to form a resultant wave. Destructive interference occurs when two waves are out of phase and cancel each other out, resulting in nodes or areas of no sound.
To find the frequency at which destructive interference will occur at the given point, we can use the formula for the path length difference (PLD) between the two speakers:
PLD = d2 - d1
where d1 is the distance from the first speaker to the point and d2 is the distance from the second speaker to the point.
In this case, d1 = 3.0 m and d2 = 3.5 m. Hence, the path length difference is:
PLD = 3.5 m - 3.0 m = 0.5 m
For destructive interference to occur, the path length difference must be an odd multiple of half the wavelength (λ/2). Therefore, we can use this relationship to find the wavelength (λ) and then calculate the frequency (f) using the speed of sound (v):
λ/2 = PLD
λ = 2 * PLD
Substituting the given values:
λ = 2 * 0.5 m = 1.0 m
Now, let's calculate the frequency:
v = λ * f
f = v / λ
The speed of sound in air is given as 343 m/s, so substituting the values:
f = 343 m/s / 1.0 m
Calculating this gives:
f = 343 Hz
Therefore, the lowest frequency at which destructive interference will occur at this point is 343 Hz.
To find the next two highest frequencies that result in destructive interference, we need to add one wavelength (λ) at a time to the initial frequency. Since λ = 1.0 m, the next two highest frequencies will have wavelengths of 2.0 m and 3.0 m.
Using the formula for frequency, we can calculate these frequencies:
f2 = 343 m/s / 2.0 m = 171.5 Hz
f3 = 343 m/s / 3.0 m = 114.3 Hz
Hence, the next two highest frequencies that result in destructive interference at this point are 171.5 Hz and 114.3 Hz.
Remember to always check your work and units to ensure accuracy in your calculations. I hope this explanation helps you understand how to approach and solve this problem!