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 6 times the fundamental frequency, which is given by:f1 = v/2L = 345 m/s / (2 * 0.25 m) = 690 HzTherefore, the sixth harmonic is:f6 = 6 * f1 = 6 * 690 Hz = 4140 HzThe second overtone of the pipe is 3 times the fundamental frequency, which is given by:f1 = v/4L = 345 m/s / (4 * 0.96 m) = 90 HzTherefore, the second overtone is:f3 = 3 * f1 = 3 * 90 Hz = 270 HzThe common resonant frequency is the lowest frequency that both the string and the pipe can resonate at, which is 270 Hz.

A 025-M String,vibrating in Its Sixth Harmonic,excites a 0