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Wize University Physics Textbook (Master) > Sound & Hearing

Constructive vs Destructive Interference

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Constructive and Destructive Interference

  • For two waves propagating from similar sources but traveling different paths, the phase would be different
  • The interference pattern between two waves depends on their phase differences
For two waves, generated by two similar in phase sources, the phase difference is equal to:

 Δφ=φ2−φ1=kx2−kx1+Δφ0=kΔx+Δφ0 \boxed{ \ \Delta\varphi=\varphi_2-\varphi_1=kx_2-kx_1+\Delta\varphi_0=k\Delta x+\Delta\varphi_0 \ }

  • kk is the wave number
  • Δx\Delta x is the path difference travelled by two waves

Conditions over the phase difference Δφ\Delta\varphi in order to have constructive or destructive interference:


Constructive Interference: Δφ=2πn ,   n=0,±1,±2, ...\Delta\varphi=2\pi n\ , \ \ \ n=0,\pm1,\pm2, \ ...










Destructive Interference: Δφ=nπ ,   n=±1,±3,±5, ... (n odd)\Delta\varphi=n\pi\ , \ \ \ n=\pm1,\pm3,\pm5, \ ... \ (n\ odd)





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Conditions over the path difference Δx\Delta x in order to have constructive or destructive interference:


Case 1: In Phase Sources:

When the sources are in phase, the waves look the same when they are emitted, and we have:

Δφ0=0   →   Δφ=kΔx=2πλ Δx\Delta\varphi_0=0 \ \ \ \to \ \ \ \Delta\varphi=k\Delta x=\dfrac{2\pi}{\lambda} \ \Delta x



Constructive Interference (in phase): Δx=mλ\Delta x=m\lambda , so we get:

Δφ=2πλ⋅mλ   →    Δφ=2πm,   m=0,±1,±2, ... \Delta\varphi=\dfrac{2\pi}{\lambda}\cdot m\lambda \ \ \ \to \ \ \ \boxed{\ \Delta\varphi=2\pi m, \ \ \ m=0,\pm1,\pm2,\ ... \ }

The waves are in phase and their interference is fully constructive.






Destructive Interference (in phase): Δx=(m+12)λ\Delta x=\bigg(m+\dfrac{1}{2}\bigg)\lambda , so we get:

Δφ=2πλ⋅(m+12)λ   →    Δφ=2π(m+12),   m=0,±1,±2, ... \Delta\varphi=\dfrac{2\pi}{\lambda}\cdot \bigg(m+\dfrac{1}{2}\bigg)\lambda \ \ \ \to \ \ \ \boxed{\ \Delta\varphi=2\pi \bigg(m+\dfrac{1}{2}\bigg), \ \ \ m=0,\pm1,\pm2,\ ... \ }


  • Alternate formula: Δx=mλ2\Delta x=\dfrac{m\lambda}{2} , from which we get  Δφ=πm ,   m=±1,±3,±5 \boxed{ \ \Delta \varphi=\pi m \ , \ \ \ m=\pm1, \pm3, \pm5 \ }


The waves are out phase and their interference is fully destructive.





Case 2: Out of Phase sources:

When the sources are out of phase, the waves look the opposite when they are emitted, and we have:

Δφ0=π\Delta\varphi_0=\pi


Constructive Interference (out of phase):  Δφ=2π(m+12),   m=0,±1,±2, ... \boxed{\ \Delta\varphi=2\pi \bigg(m+\dfrac{1}{2}\bigg), \ \ \ m=0,\pm1,\pm2,\ ... \ }










Destructive Interference (in phase):  Δφ=2πm,   m=0,±1,±2, ... \boxed{\ \Delta\varphi=2\pi m, \ \ \ m=0,\pm1,\pm2,\ ... \ }











Wize Concept
When the waves are out of phase, they are shifted by π\pi or half a wavelength with respect to each other. That's why the "nn" value is shifted by half an integer.



Exam Tip
  • Begin with the formulas for constructive interference, in phase.
  • Count how many π\bf{\pi} shifts you have (e.g. out of phase sources, reflections on rigid boundaries).
  • For each π\pi shift, swap the equations once.

Interference Shortcuts



In Phase Sources (Δϕ0=0\Delta \phi_0=0) Out of Phase Sources (Δϕ0=π\Delta \phi_0=\pi)


Constructive: Destructive:

Δx=mλ\Delta x=m\lambda Δx=mλ2\Delta x=\dfrac{m\lambda}{2}
m=0,±1,±2, ...m=0, \pm1, \pm2, \ ... m=±1,±3, ...   (odd)m= \pm1, \pm3, \ ... \ \ \ (odd)


Destructive: Constructive:

Δx=mλ2\Delta x=\dfrac{m\lambda}{2} Δx=mλ\Delta x=m\lambda
m=±1,±3, ...   (odd)m= \pm1, \pm3, \ ... \ \ \ (odd) m=0,±1,±2, ...m=0, \pm1, \pm2, \ ...


Sound Wave Interference




If two speakers with the same frequency are located a distance R1R_1 and R2R_2 from point PP, then the type of interference that occurs depends on the path difference ∣R2−R1∣\bcfi{|R_2-R_1|} between the two sound waves at that point.




For sources that are in phase:

Constructive Interference: the path difference is an integer number of wavelengths:

 ∣R2−R1∣=nλ \boxed{ \ |R_2-R_1|=n\lambda \ }
n=0,1,2,3,....n =0,1,2,3,....

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Destructive Interference: the path difference is a half-integer number of wavelengths:

 ∣R2−R1∣=(n+12)λ \boxed{ \ |R_2-R_1|=\left(n+\dfrac{1}{2}\right)\lambda \ }
n=0,1,2,3,....n =0,1,2,3,....










Wize Concept
For sources that are out of phase:
  • The pattern shifts by half a wavelength when the sources are out of phase, so the formulas also shift by half an integer.
  • This means that the equations are switched for constructive and destructive interference.




Exam Tip
  • Draw triangles: the unknown length is often the hypothenuse, so use the Pythagorean theorem to find it (a2+b2=c2a^2+b^2=c^2)
  • Use the equation v=λfv=\lambda f to switch between frequency and wavelength







Watch Out!
Read the question carefully: make sure you use the appropriate version of the equations based on whether the sources are in phase or out of phase.

Example: Speakers Out of Phase


You are standing 44 m from one of the two speakers as shown. The two speakers are out of phase. For which frequencies in the hearing range (2020 Hz to 2020 kHz) do you hear a minimum signal? (Speed of sound is 334334 m/s)



To find the path difference, we first need to find the path length d1d_1 from the top speaker to the observer:

d1=32+42=5d_1=\sqrt{3^2+4^2}=5

Now we'll find the path difference:

PD=d1−d2PD=d_1-d_2
=5−4=1=5-4=1

We have destructive interference, and the sources are out of phase, so we have to use:

PD=nλPD=n\lambda

Let's substitute PD=1PD=1 and convert from wavelength to frequency using λ=vf\lambda=\dfrac{v}{f}:

1=nvf   →  n=fv1=\dfrac{nv}{f} \ \ \ \to \ \ n=\dfrac{f}{v}

since we'll need to find the possible n\bcf n values for the given range of frequencies.


For f=20f=20 we have:

n=fv=20334=0.06n=\dfrac{f}{v}=\dfrac{20}{334}=0.06


For f=20000f=20000 we have:

n=fv=20000334=59.9n=\dfrac{f}{v}=\dfrac{20000}{334}=59.9


Therefore we could have 0.06<n<59.90.06<n<59.9, but since nn can only be an integer we get:

n=1,2,3, ... ,59n=1, 2, 3, \ ...\ , 59


Let's find the corresponding frequencies:

n=1   →   f=nv=1⋅334=334n=1 \ \ \ \to \ \ \ f=nv=1\cdot334=334 (Hz)

n=2   →   f=nv=2⋅334=668n=2 \ \ \ \to \ \ \ f=nv=2\cdot334=668 (Hz)
.
.
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n=59  →  f=nv=59⋅334=19706n=59 \ \ \to \ \ f=nv=59\cdot334=19706 (Hz)

Practice: Walking Between Speakers


Two speakers, both of frequency 410410 Hz, are placed 1.201.20 m apart. An observer, initially positioned at one of the two speakers, walks along a line perpendicular to the line joining the two speakers. Use vsound=331v_{sound}=331 m/s.

a) How far must the observer walk before reaching the first point of constructive interference with the smallest non-zero path difference?

b) How far must the observer walk before reaching the first point of destructive interference with the smallest non-zero path difference?