Question

In a location where the speed of sound is 336 m/s, a 2 000-Hz sound wave impinges on two slits 28.5 cm apart.

(a) At what angle is the first maximum of sound intensity located?

Answer 1

(a)

we have

?=d*sin?

maximum => ?=?=v/f=0.168 m

d=0.285 m

=>?=36.12⁰

(b)if ?=0.0325 m

for the same angle

d=?/sin?=0.0551 m=5.5 cm

(c)?=d*sin?

d=1?m

=>?=5.89*10^-7 m

frequency f=c/?=5.1*10¹⁴ Hz=510THz

Answer 2

we dsintheta = mlambda

sint = mlambda/d

for the first maximum m = 1

d = lambda/d

lambda = v/f

sint = (v/f)/d

sint = 20.88degree

sint = 0.589 = thetha = 36.08

b) d = lambda/sint = 5.52 cm

c) m = 1

lambda = dsint = 1 mhu x 0.589 = .589 mhum

f = v/lambda = v = c

590.33 THz