Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 cm. Find the beat frequency that they produce when playing together in their fundamentals.

Solution 41E The length of the organ pipes were 1.14 meters. Among these two, one get lengthened by 2 cm. So, it’s length is 1.14 + 0.02 = 1.16 meters. The frequency in the original pipe can be calculated as, Fn= nV / 4l. Here we are dealing with the 1st node where n=1. So, frequency of pipe 1 is, F 1 1×V = 344 = 75.43 4×1.14 4.56 The frequency of the lengthened pipe is, 1×V 344 F 2 4×1.16= 4.64= 74.13 The frequency of the beat will be, F beat= F 1 F =275.43 74.13 = 1.30 Hz . Conclusion: The beat frequency is 1.30 Hz.