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Two compact sources of sound oscillate in phase with a frequency of 100 Hz. At a point 5.00m from one source and 5.85 m from the other, the amplitude of the sound from each source separately is A. (a) What is the phase difference of the two waves at that point? (b) What is the amplitude (in terms of A) of the resultant wave at that point?

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Answer:

(a)1.557 radian (b) 1.424 A

Step-by-step explanation:

Frequency of oscillation of sound = 100 Hz


\Delta r=5.85-5=0.85m

(a)The phase difference is given by
(2\pi \Delta r)/(λ) where v is the velocity of sound in air

So phase difference
(2\pi \Delta rf)/(v) as
\lambda =(v)/(f)

So phase difference
=(2* 3.14* 0.85* 100)/(343)=1.557radian

(b) The resultant amplitude is given by
2Acos(\Phi )/(2)=2* A* cos(1.557)/(2)=1.424A

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