1)The pressure variation in a sound wave is given by P = 0.0090 sin(0.66 x 1350
ID: 1634296 • Letter: 1
Question
1)The pressure variation in a sound wave is given by P = 0.0090 sin(0.66 x 1350 t) , where P is in pascals, x in meters, and t in seconds. What is the displacement amplitude of the wave?ANSWER ITS NOT 0.0090 AND ITS NOT TAKIN 2.3X10^-13,NOR 2.4X10-14 IDK WHAT IM DOING WRONG PLEASE HELP I HAVE ONE MORE ATTEMPT LEFT. 2)Two loudspeakers are placed 3.00 m apart, as shown in the figure (Figure 1) . They emit 528 Hz sounds, in phase. A microphone is placed 3.20 m distant from a point midway between the two speakers, where an intensity maximum is recorded. A) How far must the microphone be moved to the right to find the first intensity minimum? B) Suppose the speakers are reconnected so that the 528 Hz sounds they emit are exactly out of phase. At what positions are the intensity maximum and minimum now? I GOT THE MAXIMUM WHICH IS .382M BUT I DONT KNOW HOW TO GET THE MINIMUM I THOUGHT IT WAS HALF Of THE MAXIMUM BUT ITS TELLING ME ITS WRONG
Explanation / Answer
relation between displacement maximum and pressure maximum is
Pmax = B*A*k
0.0090 = B*A*k
B is the bulk modulus = 1.42*10^5
A = ?
k = 0.66*pi
then
A = 0.009/(B*k) = 0.009/(1.42*10^5*0.66*3.142)
A = 3.05*10^-8 m
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