Question

A teacher stands well back from an outside doorway 0.85 m wide, and blows a whistle of frequency 950?Hz. Ignoring reflections, estimate at what angle(s) ?mmin (0?<?mmin<90?) it is not possible to hear the whistle clearly on the playground outside the doorway. Assume 340 m/s for the speed of sound.

Answer 1

Solution:

The given case involves the diffraction of the sound waves by the doorway of width a. We need to find the minima of sound wave intensity at certain angle(s) θ_m,min where m is the order of diffraction.

Width of the doorway, a = 0.85 m

Frequency of the sound waves, f = 950 Hz = 950/s

Speed of sound waves, v = 340 m/s

Thus the wavelength λ of sound waves is given by,

v = f*λ

λ = v/f

λ = (340 m/s)/(950/s)

λ = 0.357895 m

The m^th order minima occur at an angle of θ_m,min

a*sin(θ_m,min) = m*λ

sin(θ_m,min) = m*λ/a

Now for first order diffraction, m = 1, we have

sin(θ_1,min) = (1)*( 0.357895 m) / (0.85 m)

sin(θ_1,min) = 0.421053

θ_1,min = sin^-1(0.421053)

θ_1,min = 24.9011^o

With two significant digits,

θ_1,min = 25^o

For second order diffraction, m = 2, we have

sin(θ_2,min) = (2)*( 0.357895 m) / (0.85 m)

sin(θ_2,min) = 0.842106

θ_2,min = sin^-1(0.842106)

θ_2,min = 57.3632^o

With two significant digits,

θ_2,min = 57^o                         

For m = 3

sin(θ_3,min) = (3)*( 0.357895 m) / (0.85 m)

sin(θ_3,min) = 1.26316

One cannot take sin inverse of a number greater than 1 (or less than -1), it tells us that only two minima occur at either sides of doorway at,

θ_1,min = 25^o   and θ_2,min = 57^o

Hence the answers are

25^o, 57^o