A 40.0 cm long organ pipe is filled with air at 0°C and is open at both ends. The speed of sound in air at 0°C is 331 m/s. Another 40.0 cm long organ pipe is filled with air at 20°C and is open at both ends. What is the beat frequency between the fundamental modes of the two organ pipes?
A) 20.5 Hz
B) 17.5 Hz
C) 15.6 Hz
D) 14.9 Hz
E) 12.3 Hz

Question

Quizplus · Accepted Answer

The fundamental frequency of an open pipe with length L is given by f = nv/2L, where n is a positive integer (1, 2, 3, ...) and v is the speed of sound in air. For the first pipe at 0°C, we have L = 40.0 cm = 0.4 m and v = 331 m/s, so the fundamental frequency is f1 = (1/2L)v = 413.75 Hz. For the second pipe at 20°C, we have v' = v(1 + αΔT) where α is the coefficient of linear expansion of air (0.0034 K^-1) and ΔT = 20°C, so v' = 343.92 m/s. The fundamental frequency for this pipe is f2 = (1/2L)v' = 429.9 Hz. The beat frequency is given by |f1 - f2| = 14.9 Hz (to one decimal point). Therefore, the correct answer is D.

A 400 Cm Long Organ Pipe Is Filled with Air at at 0°C