Let us say we have 2 point sources of sound. My question is how do we consider the intensity to vary according to position? Let's say both have same amplitude, frequency and speed, just different phase.

Does intensity add up individually or do we calculate the net displacement in pressure due to the superposition and then relate max intensity to regions of max pressure?

Like here, will the intensity at A and B be the same? (Take any 2 points arbitrarily such that constructive interference is happening there) Our teacher told us this, but I'm not sure about it. enter image description here


At a point in space you have two waves arriving with amplitudes $A_1$ and $A_2$ and with wave 2 in advance of wave 1 by a phase angle of $\delta$.
I have chosen amplitudes just to be able to differentiate between the two waves.

It is not unreasonable that you add displacements if you think of one wave trying to displace a particle of the medium through which the wave is travelling by a certain amount and the other wave trying to displace the same particle by another amount.
You add those two displacements to find the resultant displacement.

Since we need to add two sinusoidal functions with the same frequency but which differ in phase, phasor addition can be used.

enter image description here

Using the cosine rule the resulting amplitude $B$ is given by $B^2 = A_1^2 + A_2^2 + 2 A_1 A_2 \cos \delta$.

Since the intensity $I$ is proportional to the amplitude squared

$I \propto A_1^2 + A_2^2 + 2 A_1 A_2 \cos \delta$.

If $A_1=A_2=A$ then $I \propto 4A^2 \cos^2 \left (\frac \delta 2\right)$ and the intensity graph is shown above.

A phase $\delta = 2 \pi$ corresponds to a path difference (pd) of one wavelength etc.

  • $\begingroup$ But what is wrong in the other fact, that intensity decreases with distance? Like the sound is louder near the speakers than far away from them regardless of any interference. $\endgroup$ – Shodai Mar 30 '16 at 7:42
  • $\begingroup$ You would need to have a factor which makes the amplitude of a wave change with distance. Usually the path lengths from the two sources are approximately the same (a few wavelengths different) so although the amplitude of the waves from each source changes they both change by approximately the same amount. $\endgroup$ – Farcher Mar 30 '16 at 8:14
  • $\begingroup$ Even if amplitude remains same, can we not say that since Power is constant, then intensity at a distance r will be Power/4*pirr since it is equally distributed in that spherical shell? $\endgroup$ – Shodai Mar 30 '16 at 8:30
  • $\begingroup$ Note that the average value of the intensity curve is $2A^2$ which is what you would expect from two sources of amplitude $A$ because when the individual intensities are added $A^2+ A^2$ you get that average intensity. The superposition of the waves from the two sources redirects the energy transfer. $\endgroup$ – Farcher Mar 30 '16 at 8:42
  • $\begingroup$ Yea but doesn't intensity reduce radially because of equi-distribution of energy? $\endgroup$ – Shodai Mar 30 '16 at 8:44

Any physics problem regarding waves, to my knowledge, will need you to compute the amplitude dependance in time and space before anything.

Waves superposition can be constructive or destructive depending on the sum of the amplitudes of the superposing waves. Intensity is defined as the product of the sound velocity by the sound pressure which is defined as the derivatice of the amplitude regarding space.

Along x axes with the example of a 1D propagating wave: $$p=\frac{\partial y}{\partial x}$$ where y is the amplitude (say $y=a\sin(\omega t -kx)$ for a plane wave), you get the instantaneous pressure: $$\textbf{I}=pc$$ where $c$ stands for the wave celerity (usually $c=\frac{\omega}{k}$). For a sum of waves you have to add $y_1$ and $y_2$ the amplitudes of both wave and do the calculation again. You'll see that intensities just don't add up.

If you want to calculate intensity over a given time $T$, you have to integrate: $$\textbf{I}_T=\frac{1}{T}\int_0^T pc\ dt$$

I hope this helps.

  • $\begingroup$ Falling short. I forgot that you could finally just end up taking the square of the sum of amplitudes since in the general case I is proportionnal to p^2. $\endgroup$ – G.Clavier Mar 29 '16 at 15:56
  • $\begingroup$ Yeah I know. This is just a TL:DR addendum for anyone not willing to do the usual calculations. Still thanks though. $\endgroup$ – G.Clavier Mar 29 '16 at 16:05

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