## Pages

`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Monday, January 14, 2008

### AP Physics B- Answer to Free Response Question on Interference

In the post dated 13th January 2008, the following free response question for practice was given to you:

Your teacher suggests to you that an interference method could be used to measure the frequency of the sound emitted by two small identical speakers S1 and S2 arranged horizontally with a separation of 0.8 m (or any suitable distance) as shown. The speakers are connected in series to a sine wave generator of constant output frequency. Your teacher moves the probe (microphone) of a sound level meter along the horizontal line AB parallel to the line S1 S2 and demonstrates to you that at O, equidistant from S1 and S2, the intensity of sound is maximum and it varies between maximum and minimum when the probe is moved along OA and OB.

(a) Explain briefly what measurements you will make to enable you to calculate the frequency of the sound emitted by the speakers.

(b) If the distance between the speakers is increased, what quantity you measure is affected and how?

(c) What equations you will use for calculating the frequency? Explain the reason for your answer to (b) based on the relevant equation.

(d) If the distance between the 2nd maximum on one side of O and the 3rd maximum on the other side of O is 1.725 m, calculate the frequency of the sound emitted by the speakers, assuming that the speed of sound at the laboratory temperature is 345 ms–1.

(e) Your teacher asks you to reverse the connection to the terminals of one of the speakers. What change, if any, will this produce in the location of the maxima and minima? Your teacher suggests to you that an interference method could be used to measure the frequency of the sound emitted by two small identical speakers S1 and S2 arranged horizontally with a separation of 0.8 m (or any suitable distance) as shown. The speakers are connected in series to a sine wave generator of constant output frequency. Your teacher moves the probe (microphone) of a sound level meter along the horizontal line AB parallel to the line S1 S2 and demonstrates to you that at O, equidistant from S1 and S2, the intensity of sound is maximum and it varies between maximum and minimum when the probe is moved along OA and OB.

(a) Explain briefly what measurements you will make to enable you to calculate the frequency of the the sound emitted by the speakers.

(b) If the distance between the speakers is increased, what quantity you measure is affected and how?

(c) What equations you will use for calculating the frequency? Explain the reason for your answer to (b) based on the relevant equation.

(d) If the distance between the 2nd maximum on one side of O and the 3rd maximum on the other side of O is 1.725 m, calculate the frequency of the sound emitted by the speakers, assuming that the speed of sound at the laboratory temperature is 345 ms–1.

(e) Your teacher asks you to reverse the connection to the terminals of one of the speakers. What change, if any, will this produce in the location of the maxima and minima? Give reason for your answer.

As promised, I give below the answer:

(a) The mean width (β) of of the interference bands is required and therefore, the distance between the centres of the nth maxima on either side of the central point O is to be measured. This distance (xn) will give the width of n bands since the point O is the centre of the zero order maximum. The value of n should be such that the line joining S1S2 to the centre of the nth band is not much different from 90º.

The distance (d) between the speakers (coherent sources) and the distance (D) between the line AB and the line S1S2 also are to be measured to facilitate the calculation of the wave length of sound.

(b) If the distance between the speakers is increased, the interference band width is decreased.

(c) The equation for band width is β = λD/d where β is the band width given by β = xn/n, λ is the wave length of the sound produced by the speakers, D is the distance between the centre of the line S1S2 and the point O and d is the separation between the speakers.

From the above equation, λ = βd/D.

The frequency f of the sound can be calculated using the equation

f = v/λ where ‘v’ is the velocity of sound.

Since the fringe width β = λD/d, when the separation between the speakers (d) is increased, β must be decreased as stated in (b).

(d) The distance between the 2rd maximum on one side of O and the third maximum on the other side of O is 1.725 m. Evidently this is the width of 5 bands since the point O is the centre of the zero order maximum. Therefore,

β = xn/n = 1.725/5 = 0.345 m so that

λ = βd/D = (0.345×0.8)/6 = 0.046 m and

f = v/λ = 345/0.046 = 7500 Hz.

(e) When the connection to the terminals of one of the speakers is reversed, the phase difference between the sound waves emitted by the speakers becomes 180º. (Earlier, they were in phase and that was why maximum intensity was observed at O). The waves therefore arrive at O with a phase difference of 180º and they interfere destructively, producing minimum intensity at O. Similarly, the intensity will be minimum at all points where it was maximum earlier. In regions where the waves arrived with opposite phase (with phase difference equal to an odd multiple of π) earlier will now become regions of maximum intensity since the waves will reach those regions in phase (with phase difference equal to an even multiple of π). Therefore, the positions of maximum intensity and minimum intensity will get interchanged.