*Question*

Two loudspeakers, A and B, are driven in phase and with the same amplitude at a frequencyÂ of 850 Hz. Point P is located 22.5 m from A and 24.3 m from B. The speed of sound is 340 m s^{â€“1}.

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(a) Deduce that a minimum intensity of sound is heard at P.Â Â [4]

(b) A microphone moves along the line from P to Q. PQ is normal to the line midwayÂ between the loudspeakers.

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The intensity of sound is detected by the microphone. Predict the variation of detectedÂ intensity as the microphone moves from P to Q.Â Â [2]

(c) When both loudspeakers are operating, the intensity of sound recorded at Q is I_{0.}Â Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound withÂ unchanged amplitude and frequency. The intensity of sound recorded at Q changes to I_{A}.

EstimateÂ Â Â Â Â [2]

(d) In another experiment, loudspeaker A is stationary and emits sound with a frequency ofÂ 850 Hz. The microphone is moving directly away from the loudspeaker with a constantÂ speed *v*. The frequency of sound recorded by the microphone is 845 Hz.Â

(i) Explain why the frequency recorded by the microphone is lower than theÂ frequency emitted by the loudspeaker.Â Â Â [2]

(ii) Calculate *v*.Â [2]

**Answer/Explanation**

**Ans: **

**a**

wavelength = \(\frac{340}{850}\) = 0.40 Â«mÂ»

path difference = 1.8 Â«mÂ» 1.8 Â«mÂ» = 4.5 Î»Â OR \(\frac{1.8}{0.20}\) =9 Â«half-wavelengthsÂ»

waves meet in antiphase Â«at PÂ»

OR destructive interference/superposition Â«at PÂ»

**b**

Â«equally spacedÂ» maxima and minimaÂ

a maximum at Q

four Â«additionalÂ» maxima Â«between P and QÂ»

**c**

the amplitude of sound at Q is halvedÂ

Â«intensity is proportional to amplitude squared henceÂ» \(\frac{I_A}{I_O}=\frac{1}{4}\)

**d i**

speed of sound relative to the microphone is lessÂ

wavelength unchanged Â«so frequency is lowerÂ»

OR

fewer waves recorded in unit time/per second Â«so frequency is lowerÂ»

**d ii** Â

845 = 850 Ã— \(\frac{340-V}{340}\)

v = 2.00 Â«m s^{-1}Â»