Standing sound waves: A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone resonance. The speed of sound in air is 345 m/s. The common resonant frequency of the string and the pipe is closest to
A) 540 Hz.
B) 360 Hz.
C) 450 Hz.
D) 630 Hz.
E) 700 Hz.

Question

Quizplus · Accepted Answer

The sixth harmonic of the string is 6f1, where f1 is the fundamental frequency of the string. The second overtone of the pipe is 3f2, where f2 is the fundamental frequency of the pipe. The common resonant frequency is the lowest frequency that both the string and the pipe can vibrate at, which is the lowest common multiple of 6f1 and 3f2. The lowest common multiple of 6 and 3 is 6, so the common resonant frequency is 6f1 = 6f2. The fundamental frequency of the string is f1 = v/2L = 345 m/s / (2 * 0.25 m) = 690 Hz. The fundamental frequency of the pipe is f2 = v/4L = 345 m/s / (4 * 0.96 m) = 90 Hz. Therefore, the common resonant frequency is 6f1 = 6 * 690 Hz = 4140 Hz. The closest answer choice is A) 540 Hz.

Standing Sound Waves: a 0