A pipe open only at one end has afundamental frequency of 256 Hz. A second pipe,

Question

A pipe open only at one end has afundamental frequency of 256 Hz. A second pipe, initiallyidentical to the first pipe, is shortened by cutting off a portionof the open end. Now, when both pipes vibrate at theirfundamental frequencies, a beat frequency of 12 hz is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is 343 m/s. A pipe open only at one end has afundamental frequency of 256 Hz. A second pipe, initiallyidentical to the first pipe, is shortened by cutting off a portionof the open end. Now, when both pipes vibrate at theirfundamental frequencies, a beat frequency of 12 hz is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is 343 m/s. A pipe open only at one end has afundamental frequency of 256 Hz. A second pipe, initiallyidentical to the first pipe, is shortened by cutting off a portionof the open end. Now, when both pipes vibrate at theirfundamental frequencies, a beat frequency of 12 hz is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is 343 m/s.

Explanation / Answer

The equation to find beat frequency is fbeat =f2 - f1 256 + 12 = 268Hz (it must be higher since the length is nowshorter) Now to find how much was cut off, you need to see how long theinitial one is and how long the second one is. The equation that you use is f = v/4L for a tube closed at oneend's fundamental frequency. L = v/4*f = (343m/s)/4*256Hz = 0.335m L = v/4*f = (343m/s)/4*268Hz = 0.32m So 0.335m - 0.32m = 0.015m were removed Hope that helps

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