# Frequency at which no destructive interference occurs

• prodo123
In summary: Since x(n) increases as n→1, I think n=1 is the first point of destructive interference from ∞→x. Therefore if x(1)=0, there are no other point of destructive interference other than the source B itself.x(0) as well as x(1) equals 0 at ƒ=86 Hz. What does this mean?It means that at ƒ=86 Hz, both speakers are producing the same amount of sound, and there is no difference between x(0) and x(1).
prodo123

## Homework Statement

Two speakers A and B, 2.00 m apart, produce a sine wave at the same frequency and phase. A microphone is placed on the line BC perpendicular to AB, at a distance x from B. The speed of sound is v=344 m/s.

For a frequency ƒ low enough, there will be no destructive interference along the line BC. Find this frequency.

## Homework Equations

Let ##r=\sqrt{x^2+2^2}## be the distance from speaker A to the point along the line BC.

##y_{A}(x,t)=A\cos(kr-\omega t)\\
y_{B}(x,t)=A\cos(kx-\omega t)\\
k=\frac{2\pi f}{v}\\
\omega = 2πf##

## The Attempt at a Solution

Destructive interference occurs when ##y_{A} = -y_{B}##:

##A\cos(kr-\omega t) = -A\cos(kx-\omega t)\\
A\cos(kr-\omega t) = A\cos(kx-\omega t+(2n-1)\pi)\\
kr = kx+(2n-1)\pi\\
r=\sqrt{x^{2}+2^{2}}=x+\frac{(2n-1)\pi}{k}\\
r=x+\frac{(2n-1)v}{2f}##

For simplicity, let ##C = \frac{(2n-1)v}{2f}##

##r^2 = x^2+4 = x^2+2cx+c^2\\
2cx = 4-c^2\\
x(n) = 2/c - c/2\\
x(n) = \frac{4f}{(2n-1)v} - \frac{(2n-1)v}{4f}##

A previous part of the problem set ƒ = 786 Hz and asked to find points of destructive interference; values for which x(n) > 0 found points of destructive interference at n=1,2,3,4,5, which correlated with the answer in the back.

Since x(n) increased as n→1, I think n=1 is the first point of destructive interference from ∞→x. Therefore if x(1)=0, there are no other point of destructive interference other than the source B itself.

Finding the frequency ##f## for which ##x(1)=0##:

##x(1) = \frac{4f}{v}-\frac{v}{4f}= 0\\
4f=v\\
f=\frac{v}{4} = 344/4 = 86\text{ Hz}##

which is the correct answer in the textbook.I'm having trouble interpreting the equation x(n) that I derived.

1. There are also negative values of x(n) for ##n>5## and ##-4<n\le 0##, and positive values for ##n\le -5## when ƒ=786 Hz. Since x is a point along the continuous line BC, and r is defined for all x, why doesn't the textbook count all nonzero values of x(n) as points of destructive interference? Or if it's looking specifically for ##x>0##, why doesn't it count x(n) for ##n<1##?

2. If x(1) = 0 then isn't B itself a point of destructive interference?

3. x(0) as well as x(1) equals 0 at ƒ=86 Hz. What does this mean?

4. Is my interpretation correct that if the only point of destructive interference is x(1) = 0, speaker B is not able to output any sound since the source itself is destructively interfered?

#### Attachments

• IMG_7268.JPG
38.3 KB · Views: 808
Last edited:
berkeman
Hello Prodo,

1.) There is indeed symmetry wrt the line AB. I don't have the exact problem statement for
prodo123 said:
A previous part of the problem
however, if it says 'distance from B' then all is well.

2. It sure is.

3. It means that the distance from B is zero when you are at B. That is not what you mean, I suppose . You mean to say something about ##y_A+y_B##, but your equation for ##y## does not take into account that ##A## also varies with distance.

4. Therefore your interpretation is not correct.

BvU said:
Hello Prodo,

1.) There is indeed symmetry wrt the line AB. I don't have the exact problem statement for
however, if it says 'distance from B' then all is well.

2. It sure is.

3. It means that the distance from B is zero when you are at B. That is not what you mean, I suppose . You mean to say something about ##y_A+y_B##, but your equation for ##y## does not take into account that ##A## also varies with distance.

4. Therefore your interpretation is not correct.

Thanks for the reply. The chapter doesn't handle displacement amplitude decay over distance, which is why I held A constant.

The issue is that there's supposed to be reciprocity but the textbook ignores (or denies?) it. The previous part of the question simply says:
At what distances from B will there be destructive interference?
Values of ##x(n)## for n=1,2,3,4,5... give 9.01m, 2.71m, 1.27m, 0.534m, 0.026m which are the five values given as the answer to the problem. I'll give the textbook the benefit of doubt of looking for positive ##x## only, since the diagram shows BC going in the positive-x direction only.
But even so ##x(-5)## results in 0.372m, and ##x(n)## for all ##n\le -5## is positive. Looking at the graph of ##y_A+y_B## these are indeed points where the two waves cancel out.
I'm honestly starting to think the textbook got lazy and didn't bother to write down all the solutions...

prodo123 said:
The chapter doesn't handle displacement amplitude decay over distance
But you know from experience.

prodo123 said:
there's supposed to be reciprocity
reciprocity ?

At what distances from B will there be destructive interference?
It does not ask for ##x##, it asks for distances. Distances are alsways positive. They can be to the left or to the right. If this is with 786 Hz, then I don't understand the given answers. Can you explain ?

BvU said:
But you know from experience.
Yes, I know from experience but that's not what the question is asking for and including it will give the wrong answer to the problem.

BvU said:
reciprocity ?

It does not ask for ##x##, it asks for distances. Distances are alsways positive. They can be to the left or to the right. If this is with 786 Hz, then I don't understand the given answers. Can you explain ?

Sorry, meant to say symmetry across ##x=0##.

After some discussion and playing around with the graph of ##y_A+y_B## I concluded on the following:
• Diagram specifically points to the right only. Mathematically negative ##x## exists but the problem ignores that.
• The textbook is looking for positive distances from ##B## at which total cancellation occurs for all t.
• The given answers for 786 Hz (9.01m, 2.71m, 1.27m, 0.534m, 0.026m) are the only points on BC where this occurs under the given conditions and ##f=786\text{ Hz}##.
• Graph of ##x(n)## (treating ##n## as continuous instead of an integer) shows it's a hyperbola, and the textbook's answers correspond to the positive values of the right half of the hyperbola only.
• If ##f=86\text{ Hz}##, ##x=0## is the only place this occurs.
• Apparently ##x\ne 0## (the speaker B itself is not a part of BC?), therefore the question considers ##f=86\text{ Hz}## to have no such points on BC even though it occurs at ##x=0##.
The solutions for ##x(n)## outside ##[1,...,5]## are zeroes (total cancellation) of ##y_A+y_B## at time ##t=0##, but do not hold at other values of ##t##. For example, ##x(-5)=0.372## is not a zero at time ##t=10##.

I've decided it's just a really poorly worded question and moved on.

berkeman
prodo123 said:
##x(n)## for n=1,2,3,4,5... give 9.01m, 2.71m, 1.27m, 0.534m, 0.026m
I find these values if I take ##f = 784## Hz, not 786 ?!
prodo123 said:
Diagram specifically points to the right only. Mathematically negative ##x## exists but the problem ignores that
No. ##x## is a distance, a length. And a length is non-negative.

You wanted to solve ##r=\sqrt{x^{2}+2^{2}}=x+\frac{(2n-1)\pi}{k}## and in order to do so, you squared left and right. Did it occur to you that you introduced wrong answers that way ? Did you check that your ##x(-5) = -x(6)## satisfies the original equation (it does not) ?

prodo123
BvU said:
I find these values if I take ##f = 784## Hz, not 786 ?!
No. ##x## is a distance, a length. And a length is non-negative.

You wanted to solve ##r=\sqrt{x^{2}+2^{2}}=x+\frac{(2n-1)\pi}{k}## and in order to do so, you squared left and right. Did it occur to you that you introduced wrong answers that way ? Did you check that your ##x(-5) = -x(6)## satisfies the original equation (it does not) ?

Sorry, I got confused, it's for 784 Hz.

Showing that ##x(-5)## is indeed equal to ##-x(6)##:

##x(-5)=\frac{4*784}{344(2(-5)-1)}-\frac{344(2(-5)-1)}{4*784}\\
x(-5)=\frac{3136}{-3784}-\frac{-3784}{3136}\\
x(6)=\frac{4*784}{344(2(6)-1)}-\frac{344(2(6)-1)}{4*784}\\
x(6)=\frac{3136}{3784}-\frac{3784}{3136}\\
-x(6)=\frac{-3136}{3784}-\frac{-3784}{3136}=x(-5)\\
\\
x(-5)=\frac{70065}{185416}\approx 0.37788##

as for ##y_A=-y_B## at time ##t=0##:

##y_A+y_B=0\\
A\cos(\frac{784*2\pi}{344}x(-5))+A\cos(\frac{784*2\pi}{344}\sqrt{x(-5)^2+4})=0\\
\cos(\frac{70065\pi}{40678})+\cos(\frac{377393\pi}{40678})=0\\
0.64329-0.64329=0\\
0=0##

Graph of ##y_A+y_B## showing that 0.37788 is indeed a root:

#### Attachments

• Screen Shot 2018-04-21 at 1.48.16 AM.png
37.9 KB · Views: 528
Here's a comparison of the roots for ##y(x)=y_A+y_B, f=784\text{ Hz}##, assuming an arbitrary amplitude of 1 for both waves, found at different values of ##t##.

The given set of roots include ##x(4)=0.026## and ##x(5)=0.534## (blue); the root in question is ##x(-5)=0.37788## (red).

at ##t=0##:

at ##t=0.8##:

at ##t=1.3##:

which shows ##x=0.37788## to be a root only when ##t=0##, while the given answers hold true for all ##t##.

BvU said:
You wanted to solve ##r=\sqrt{x^{2}+2^{2}}=x+\frac{(2n-1)\pi}{k}## and in order to do so, you squared left and right. Did it occur to you that you introduced wrong answers that way ?

I think you're right about the squaring introducing false answers, but then how would one prove that a root at some time ##t## is also a root for all ##t##?

#### Attachments

• Screen Shot 2018-04-21 at 2.02.00 AM.png
38.4 KB · Views: 540
• Screen Shot 2018-04-21 at 2.01.47 AM.png
37.4 KB · Views: 526
• Screen Shot 2018-04-21 at 2.05.22 AM.png
33.1 KB · Views: 539

## 1. What is the concept of destructive interference?

Destructive interference occurs when two waves of equal frequency and amplitude meet and cancel each other out, resulting in no net displacement of the medium.

## 2. How does destructive interference affect the frequency of a wave?

The frequency of a wave remains unchanged during destructive interference. It is the amplitude that is affected, as it decreases due to the waves cancelling each other out.

## 3. Can destructive interference occur at any frequency?

No, destructive interference can only occur at specific frequencies where the waves have opposite phase and can cancel each other out. These frequencies are known as nodes.

## 4. What happens when there is no destructive interference?

When there is no destructive interference, the waves will superimpose and create a resultant wave with an increased amplitude. This is known as constructive interference.

## 5. How is the frequency at which no destructive interference occurs determined?

The frequency at which no destructive interference occurs can be determined by finding the difference in path length between the two waves and dividing it by the wavelength of the waves. This will give the number of wavelengths that the waves are out of phase, and the frequency can be calculated from there.

• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
999
• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
274
• Introductory Physics Homework Help
Replies
10
Views
337
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
458
• Introductory Physics Homework Help
Replies
17
Views
511
• Introductory Physics Homework Help
Replies
8
Views
646