#### Question

# CIE Physics - waves and path difference please help?

Two sources S1 and S2 of sound are situated 80 cm apart in air. The frequency of vibration can be varied. The two sources always vibrate in phase but have different amplitudes of vibration.

A microphone M is situated a distance 100 cm from S1 along a line that is normal to S1S2.

As the frequency of S1 and S2 is gradually increased, the microphone M detects maxima and minima of intensity of sound.

The speed of sound in air is 330 m s–1.

The frequency of the sound from S1 and S2 is increased.

Determine the number of minima that will be detected at M as the frequency is increased from 1.0 kHz to 4.0 kHz.

I don't understand this question or how to answer it at all. Please could someone explain to me in detail so that I will be able to answer questions like this one.

The question is worth four marks and the four steps shown in the answer booklet are:

Path difference between waves from S1 and S2 = 28 cm.

Wavelength changes from 33 cm to 8.25 cm.

Minimum when λ = (56 cm,) 18.7 cm, 11.2 cm, (8.0 cm).

So there are two minima.

Thanks so much for explaining to me... I really appreciate it!

#### Answer

Poor quality explanation in the booklet, isn't it?

Up to the point where the author states: "Wavelength changes from 33 cm to 8.25 cm.", it's fine.

But he then takes a 'quantum leap' and blandly states: "Minimum when λ = (56 cm,) 18.7 cm, 11.2 cm, (8.0 cm)." - without any justification whatsoever as to how he reached this conclusion!

No wonder you're confused; the author of the booklet is clearly NOT a physicist nor a mathematician!

---------------------------

Thank you for your comment. It's caused me to have a good, long think about this: here's my reasoning . . .

Use a pencil and paper & draw some sine-waves, shifted apart in time. Now think about the phase relationships between the waves arriving at the microphone from the two sources. It can be seen that when the path difference from the two sources is n.(w/2), there will be constructive OR destructive interference at the microphone, where w = the wavelength of the sound and n is an integer,

If n is an odd integer, there will be destructive interference.

If n is an even integer, there will be constructive interference.

(It's the destructive interference that is being asked for in the question).

Now, a bit of elementary math is required . . .

Using Phythagoras Theorem (right-angled triangles, etc.),

we can calculate that the difference in the distance from the most distant sound source to the microphone to that of the nearest sound source is 0.28 metre.

So, n.(w/2) = the path difference = 0.28 (metres) . . . . [1]

for constructive or destructive interference (depending on whether n is odd or even).

Now since speed of sound in air, v = (frequency of the sound, f) times (its wavelength, w),

we get: wavelength = speed / frequency,

that is: w = v / f

{That lot above is the key 'Physics' part of this Q.}

Now we know v; it is 300 m/s.

From [1] above, we can now say:

n.(300/2f) = the path difference = 0.28.

This can easily be re-arranged to give:

300n = 0.56f, which simplifies to:

f = 535.7n . . . . . .[2]

Now for the key 'Maths' bit . . . .

We know that if n = an even number, there will be constructive interference.

If you put successive even numbers into [2], you get the following frequencies:

n = 0; f = 0

n = 2; f = 1070;

n = 4; f = 2143;

n = 6; f = 3215;

n = 8; f = 4286

. . .etc . . .

And:

We know that if n = an odd number, there will be destructive interference.

If you put successive odd numbers into [2], you get the following frequencies:

n = 1; f = 536

n = 3; f = 1607;

n = 5; f = 2679;

n = 7; f = 3750;

n = 9; f = 4822

. . .etc . . .

All the freqs. are approximate - but I'm sure you'll follow the reasoning . . . .

Now we require those frequencies that are in the range from 1 to 4000 Hz. Looking at the destructive interference table above, we can see that there are THREE * frequencies that lie in this range: 1607, 2679 and 3750 Hz.

Calculating their corresponding wavelengths is trivial, since we know that wavelength = speed / frequency.

Hence, the corresponding wavelengths are :

19 cm.; 11 cm. and 8 cm. (approximately).

* Note: At this point, I disagee with the 'published' answer! (It says "2 minima"). And that was why this Q. took me so long to reply! I had to be really sure that my reasoning was bullet-proof!

O.K. - not the easiest Q. to answer - but you did ask for a "detailed reply". . . . I hope that lot was clear enough!

Skywave's method is correct, only reason for the wrong answers is the usage of 300m/s instead of 330m/s.

THANKS SKY!