Wave Interference - Calculate Possible Frequencies | Best Buy

In summary, a student conducts an experiment at Best Buy with a pair of speakers, placing them 4.0 m apart and producing a single tone using a signal generator. When the student stands at a point 2.0 m from one speaker and 2.3 m from the other, a quiet spot is noticed. Knowing the speed of sound in the room (350 m/s), the possible frequencies played by the speakers can be calculated using the equation f = v/λ = 350/6(2n + 1), where n is an integer. The quiet spot is created due to destructive interference, with the waves being out of phase by 1/2 of a wavelength. This results in potential frequencies of 583
  • #1
Arnoldjavs3
191
3

Homework Statement


A student enters Best Buy prepared to buy a pair of speakers. Before he does so, he conducts an experiment with them. He places the speakers 4.0 m apart and connects a signal generator to both speakers that produces a single and consistent tone. (constant wavelength and frequency) He then walks to a point that is 2.0 m from one speaker and2.3 m from the other. At that point he notices a quiet “spot”. If the speed of the sound in the room is known to be 350 m/s, calculate the possible frequencies being played by the speakers.

Homework Equations



PD = (n-1/2) λ

The Attempt at a Solution


f = v/(lambda) = 350/6(2n + 1) = 58.33/(2n+1) ... n belongs to integer
Sorry i have no idea what I'm doing nor do i understand the question...
Could someone please direct me to a couple of videos/helpful links that would actually help me understand what's going on here? This is for my online physics course that I'm taking, and it's hard to actually follow along through a webpage.
 
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  • #2
Make drawing of the set up showing the distances.

Do you know what they mean by "a quiet spot" ?
 
  • #3
I'll be honest this course is osmething I haven't touched for a while as I've been busy with my actual school. I'm rusty on this subject.

Pardon my ignorance where necessary, but I believe a quiet spot indicated that the area had a destructive interference?

Do you know of any reliable youtube channels that I can rely on?
 
  • #4
Pardon my ignorance where necessary, but I believe a quiet spot indicated that the area had a destructive interference?

That's correct.

Try making the drawing and work out what must happen to get destructive interference at that point. Remember that both speakers are connected to the same signal generator so both speakers produce waves with the same wavelength.

Try this youtube clip...

 
  • #5
hi; i just wanted to thank you for the help you have given me.

That video was helpful. But I am stuck here, as i know the extra distance is .3m, and i know that they are out of phase and because of this the quiet spot is created. What i don't know is the wavelength nor the frequency. I understand that they are the same because of the signal generator, but where do I go from here? How can i find(and confirm) what the potential frequencies are?
 
  • #6
Okay; Just to ask. Since i know it is .3m apart, can i use submultiples of this to formulate information for the wavelength?
E.g, .1 m ->

frequency = 350/.1 ->3500 Hz
 
  • #7
Arnoldjavs3 said:
What i don't know is the wavelength nor the frequency.
But you do know the product of them.

Arnoldjavs3 said:
i know the extra distance is .3m, and i know that they are out of phase and because of this the quiet spot is created.
By what fraction of a wavelength can the waves be out of phase, in order to create a quiet spot?
 
  • #8
1/2 of a wavelength ? So they can be out of phase.

I ended up getting 583.33 Hz as a potential frequency? I used the speed of sound in the room and divided it by .6 as that is the wavelength
 
  • #9
Arnoldjavs3 said:
I ended up getting 583.33 Hz as a potential frequency? I used the speed of sound in the room and divided it by .6 as that is the wavelength
Right.

Arnoldjavs3 said:
1/2 of a wavelength ?
That is one possibility; are there any others?
 
  • #10
I believe that any multiple of .5 will create a deconstructive interference?(.5, 1.5, 2.5).

Okay! I think I got it now. THank you so much for your help!
 

What is wave interference?

Wave interference is the phenomenon that occurs when two or more waves meet at the same point in space. The waves will combine and either increase or decrease in amplitude, depending on their relative phase.

How is wave interference calculated?

The calculation of wave interference involves determining the phase difference between the waves, the amplitudes of the waves, and the relative positions of the waves. This information is used to calculate the resulting amplitude of the combined waves.

What are the possible frequencies that can be produced by wave interference?

The possible frequencies that can be produced by wave interference depend on the frequencies of the individual waves involved. The resulting frequency will be a combination of the frequencies of the original waves, and can range from the sum or difference of the two frequencies, to a completely new frequency.

How does wave interference affect sound and light waves?

Wave interference affects both sound and light waves in similar ways. When waves interfere constructively, the resulting amplitude will be greater than that of the original waves. When waves interfere destructively, the resulting amplitude will be less than that of the original waves. This can lead to the amplification or cancellation of sound and light.

How can wave interference be observed or utilized in everyday life?

Wave interference can be observed in many everyday situations, such as when listening to music through speakers, or seeing rainbows in the sky. It is also utilized in many technologies, such as radio and television signals, as well as in medical imaging techniques like MRI scans.

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