Interferenza di onde armoniche en

From "Physics, Waves, Music": a website about the physics of waves and sound, acoustics of musical instruments, musical scales, harmony and music.

AI-translated from Italian. Original page: Interferenza_di_onde_armoniche

Interference of harmonic waves

Imagine two point sources S1 and S2 that emit periodic waves with the same amplitude, frequency and phase.

Imagine also that the oscillatory motion of the two point sources S1 and S2 is "harmonic", that is, expressed by a mathematical function of the type

yS1(t)=yS2(t)=Asin⁡(ωt+ϕ)

(for a quick introduction to trigonometric functions see the page the sine and cosine functions)

symbol meaning
A amplitude of the sinusoid
ω angular frequency. It is related to the period T by the expression ω=2πT
ϕ phase of the wave. It specifies the initial position of the source at time t=0. Indeed yS(0)=Asin⁡(ϕ)

These waves propagate in the space surrounding the sources and, when they reach the same point P, they superpose and interfere. We ask under what conditions point P is a point of (totally) constructive interference. It is easy to see that this happens when two "maxima", so to speak, arrive at point P at the same instant.


The conditions for constructive and destructive interference at a generic point P obtained for generic periodic waves, already expressed in terms of differences of times and lengths on the page mathematical aspects of interference, can now be derived by exploiting the particular mathematical properties of the sine functions that superpose at point P.

Calling t1 and t2 the times the wave takes to travel the distance separating P from S1 and S2, we see that at point P the motion of the sources is reproduced with a certain delay, given precisely by t1 and t2 (see the section how do we describe a wave?).

The total wave function at point P is therefore given by

y(𝐏,t)=yS1(t−t1)+yS2(t−t2)
=Asin⁡[ω(t−t1)+ϕ]+Asin⁡[ω(t−t2)+ϕ]
=Asin⁡θ1(t)+Asin⁡θ2(t)

where we have set

θ1(t)=ω(t−t1)+ϕ
θ2(t)=ω(t−t2)+ϕ

Constructive interference of harmonic waves in phase

The condition for constructive interference is obtained simply by requiring that the difference (in absolute value) of the arguments θ1(t) and θ2(t) be a multiple of 2π. We then obtain

|θ1(t)−θ2(t)|=ω|t1−t2|=2kπ with k=0,1,2,… (1)


From (1) we deduce that the condition for constructive interference is independent of the time t and depends only on the position of point P (uniquely determined by the delay times t1 and t2). Moreover, since ω=2πT, (1) becomes

ω|t1−t2|=2πT|t1−t2|=2kπ

from which

|t1−t2|=k⋅T with
k=0,1,2,… (1*)

which follows the condition in terms of time already derived in the general conditions for interference.

Since t1=d1/v and t2=d2/v, where d1 and d2 are the distances of P from the two sources and v is the speed of the wave, and since vT=λ, the same condition can be written in terms of path difference:

|d1−d2|=kλ

that is, interference is constructive at the points whose distances from the two sources differ by a whole number of wavelengths.

Destructive interference of harmonic waves in phase

The condition for destructive interference is obtained simply by requiring that the difference (in absolute value) of the arguments θ1(t) and θ2(t) be an odd multiple of π. We then obtain

|θ1(t)−θ2(t)|=ω|t1−t2|=(2k+1)π with
k=0,1,2,… (2)

From (2) we deduce that the condition for destructive interference is independent of the time t and depends only on the position of point P (uniquely determined by the delay times t1 and t2). Moreover, since ω=2πT, (2) becomes

ω|t1−t2|=2πT|t1−t2|=(2k+1)π

from which

|t1−t2|=(2k+1)⋅T2 with k=0,1,2,…, (2*)

which follows the condition in terms of time already derived in the general conditions for interference. In terms of path difference:

|d1−d2|=(2k+1)λ2

that is, interference is destructive at the points whose distances from the two sources differ by an odd number of half wavelengths.

Interference of out-of-phase harmonic waves

Imagine that the two point sources S1 and S2 emit harmonic waves with the same amplitude and frequency, but not synchronised. Assume for example that source S2 reproduces the oscillation of source S1 with a certain time shift, which we will denote by ts. We will take

ts>0 if source S2 oscillates later than S1
ts<0 if source S2 oscillates earlier than S1

If the motion of source S1 is described by the function

yS1(t)=Asin⁡(ωt+ϕ),

then the motion of source S2 is described by

yS2(t)=yS1(t−ts)=Asin⁡(ω(t−ts)+ϕ).

If, moreover, the time shift ts is a known fraction of the period T, that is

ts=±αT with α∈]0;1[

recalling, as usual, that ω=2πT, we obtain

yS2(t)=Asin⁡[ω(t∓αT)+ϕ]=Asin⁡(ωt+ϕ∓2πα).

In practice the time shift introduces a phase angle 2πα, which allows the conditions for constructive (1) and destructive (2) interference to be reformulated at once.

Setting

θ1(t)=ω(t−t1)+ϕ
θ2(t)=ω(t−t2)+ϕ∓2πα

condition (1) becomes

|θ1(t)−θ2(t)|=|ω(t1−t2)±2πα|=2kπ with
k=0,1,2,… (1**)

and condition (2) becomes

|θ1(t)−θ2(t)|=|ω(t1−t2)±2πα|=(2k+1)π with
k=0,1,2,… (2**)

Interference between sources of different frequencies

All the previous interference conditions were derived under the assumption that the two interfering sources had the same frequency. The resulting conditions are independent of time and depend only on the position of point P in space.

If the sources do not oscillate at the same frequency, the interference conditions at a given point in space evolve in time, which makes the study of the phenomenon extremely complex. Suppose we have two "harmonic" sources S1 and S2 (with the same amplitude and phase) of different frequency. The oscillatory motion of the two sources can, as usual, be expressed by

yS1(t)=Asin⁡(ω1t+ϕ),
yS2(t)=Asin⁡(ω2t+ϕ),

with ω1≠ω2; repeating the derivation carried out, for example, in the case of constructive interference, condition (1) now becomes

|θ1(t)−θ2(t)|=|(ω1−ω2)t−ω1t1+ω2t2|=2kπ with
k=0,1,2,… (1***)

which, as can be seen, depends on the time t.

Beats

If the two sources of different frequencies are located at the same point in space, it is clear that, for a fixed point P, the delay times t1 and t2 coincide. If we set t1=t2=xv, the interference condition (1***) remains time-dependent, but takes the simpler form:

|θ1(t)−θ2(t)|=|(ω1−ω2)(t−xv)|=2kπ with
k=0,1,2,… (1°)

which can be written as:

|t−xv|=2kπ|(ω1−ω2)|=kT1⋅T2|(T1−T2)| with
k=0,1,2,… (1°°)

where, as usual, T1 and T2 denote the periods of the two sources. This condition gives all the instants of time t at which, at point P, at distance x from the two sources, there is constructive interference. If we consider two consecutive instants (that is, taken for values k and k+1) and evaluate their difference Δt

Δt=T1⋅T2|(T1−T2)|

we realise that, for a fixed point in space, the interference condition recurs every T1⋅T2|(T1−T2)| seconds, that is, it behaves periodically with frequency

f=1Δt=|(T1−T2)|T1⋅T2=|f1−f2|

where, in the last step, the general relation between period and frequency was used, setting f1=1T1 and f2=1T2.

The frequency f=|f1−f2| is called the beat frequency. By taking f1 and f2 almost equal, the beat frequency can be made much lower than the individual frequencies of the two interfering sources.

One last remark: the whole argument started from the condition for constructive interference. It remains valid, however, if one starts from condition (2) for destructive interference.

Check your understanding

Problem. Two loudspeakers emit in phase the same sound, with a wavelength of 50 cm. A listener is 3.00 m from the first and 3.75 m from the second. What kind of interference do they perceive? Solution. The path difference is 0.75 m, that is 1.5 wavelengths: an odd number (3) of half wavelengths. The interference is destructive (k = 1 in the condition for d1−d2) and the sound is weak.

Problem. In the situation of the previous problem, what is the lowest frequency for which there is destructive interference at the listener's position? (speed of sound 340 m/s) Solution. The condition with k = 0 requires the path difference, 0.75 m, to be half a wavelength: λ = 1.5 m, that is f = 340/1.5 Hz ≈ 227 Hz. The next frequencies with destructive interference are 3 times, 5 times, … this one: 680 Hz (λ = 50 cm, the case of the previous problem), 1133 Hz, …

Problem. If the second loudspeaker is connected with its wires swapped, it oscillates in antiphase with respect to the first (α=1/2). What is heard at a point equidistant from the two loudspeakers? Solution. The phase shift adds 2πα=π to the phase difference. At the equidistant point t1=t2, so the phase difference is π and the interference is destructive: the conditions for constructive and destructive interference are swapped with respect to the case of sources in phase.

Problem. Two tuning forks vibrate at 440 Hz and 443 Hz. How often does constructive interference repeat at a point near the two tuning forks? Solution. The beat frequency is

Virtual laboratory: interference experiments

If you are interested in visualising interference phenomena, you can use our 2D Waves virtual laboratory. Below we propose a series of guided experiments, but you are free to try new configurations and new experiments. The app is very intuitive; to understand correctly what you observe, just keep in mind that the coloured areas represent the amplitude of the oscillation. The brighter the colour, the greater the amplitude of the oscillations (green for crests, red for troughs). Obviously the dark areas represent the absence of oscillation caused by destructive interference.

Suggested experiments:


"Physics, Waves, Music": a website about the physics of waves, acoustics of musical instruments, musical scales, harmony and music.