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Chapter 17, Problem 17 A 3.24-kHz tone is being produced by a speaker with a diameter of 0.202 m. The air temperature changes from 0 to 23 °C Assuming air to be an ideal gas, find the change in the diffraction angle θ. (colder) θ2(hotter) Number Units the tolerance is +/-290
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For an ideal gas, 71 And ykT 71 From the above two equations, The speed of the sound at 0°C is 331 m/s Hence, (31 m/)273+0 273 +23 344.66 m/s For the circular speaker the condition for the diffraction is, 1.222 Hence, the change in the diffraction angle is as follows 1.22 11.22 Since λ 1.221 1.22 (344.66m/s) 3.24 10Hz0202m82410 H0.202m) S11.22(331m/s) - Sin -Sin 3.24x103Hz (0.202m - 39.976-38.098 =1.878

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