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...

Question

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

Answer 1

see question 48 as an example

http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PST4_2/T4_2.htm

Answer 2

everything made sense I just don't see why .5 m must be half a wave length

Answer 3

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!

Answer 4

please if someone else knows...

Answer 5

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.

Answer 6

idk !!!! im having trouble on the same question !!!! GOT NOO IDEA !!