Question

Two in-phase loudspeakers are located at (x,y)coordinates (?3.0m,+2.0m) and (?3.0m,?2.0m). They emit identical sound waves with a 2.0 m wavelength and amplitude a.

Determine the amplitude of the sound at the five positions on the y-axis (x=0)

a)with y=0.5m

b)with y=1.0m

c)with y=1.5m

d)with y=2.0m

Answer 1

Two the wavefronts of 2 sound waves having same freq, wavelength and amplitude, and which do not have any intial phase diff, advance then at various points of superposition, the resultant amplitude depends on the path difference (p) of interfering wavelets. p is related to phase difference (d) by wave vector (k = 2pi/lamda)

phase diff = (k) path diff
d = (2pi/lamda) * p
-------------------------------------
in wave equation y = A sin (kx - wt +phi)
y = A sink [k(x-vt) - phi] >>>> look inside sine (x-vt) path diff
--------------------------------------...
the amplidute of resultant wave of 2 waves interfering [y1= a1 sin wt and y2 = a2 sin wt (every thing same, but let us different amplitudes - to show maxima minima)] is given by

A^2 = a1^2 + a2^2 + 2 a1*a2 cos [ phase difference (d) ]

this d is caused by path difference on projected screen at desired place.
now a1=a2=a in your case

A^2 = 2a^2 {1 + cos [d ]}
A^2 = 2a^2 {1 + cos [2i p / lamda]}
--------------------------------------
for a point x=0 , y=0
path difference p = r 1 - r 2 =0, also d=0
wavelets reach this point with same path or in step of phase

A^2 = 2a^2 {1 + cos [0]} = 4 a^2
A = 2a (twice of each wave amplitude) = a+a added for maxima. Wherever you keep your screen/ear say x =any distance as long as y=0, you will hear maximum sound on dividing bisector
--------------------------------------...
at x =0 , y=0.5
p = r1 - r2
r1 = sqrt [2^2 +2.5^2] = sqrt [10.25] = 3.20
r2 = sqrt [2^2 +1.5^2] =sqrt [6.25] = 2.50
path diff p = 0.7
phase diffr d = (2*3.14/2.4)*0.7 = 1.8316= 105 degree

cos (105 deg) = - 0.2578

A^2 = 2a^2[1 - 0.2578] = 1.4823 a^2
A = 1.217 a