A 3.00 kHz tone is being produced by a speaker with a diameter of 0.175 m. The air temperature changes from 0 to 29°C. Assuming air to be an ideal gas, find the change in the diffraction angle

My solution

v directly proportional to sqrt T

v = sqrt 302 =17.382

lambda = v/f = 17.382/3000
=0.005794

Sin theta = 1.22 lambda/ diameter

= 1.22x 0.005794/0.175
=0.009425
theta = Sin^-1 0.009425
= 0.54

answer is wrong

1 answer

We are given the initial temperature T1 = 0°C and the final temperature T2 = 29°C. The speed of sound in air changes with temperature as follows:

v(T) = v0 * sqrt(T/T0)

where v0 is the speed of sound at a reference temperature T0. We can use the speed of sound at 0°C as our reference, v0 = 331.5 m/s (according to the standard model of the atmosphere).

The initial speed of sound (at 0°C) is v1 = v0 = 331.5 m/s. The final speed of sound (at 29°C) is:

v2 = v0 * sqrt((T2 + 273) / (T1 + 273))
v2 = 331.5 * sqrt((29 + 273) / 273)
v2 ≈ 347.6 m/s

Now we can compute the initial and final wavelengths:

lambda1 = v1 / f = 331.5 / 3000 ≈ 0.1105 m
lambda2 = v2 / f = 347.6 / 3000 ≈ 0.1159 m

The diffraction angle can be found using the formula:

sin(theta) = 1.22 * lambda / diameter

First, let's compute the initial angle:

sin(theta1) = 1.22 * 0.1105 / 0.175
sin(theta1) ≈ 0.7709
theta1 = arcsin(0.7709) ≈ 50.4°

Now let's compute the final angle:

sin(theta2) = 1.22 * 0.1159 / 0.175
sin(theta2) ≈ 0.8079
theta2 = arcsin(0.8079) ≈ 53.6°

Finally, the change in the diffraction angle is:

delta_theta = theta2 - theta1 ≈ 53.6 - 50.4 ≈ 3.2°