Two identical closed-open pipes each have a fundamental frequency of 500 Hz.What percent change in the length of one of the tubes will cause a beat frequency of 10 Hz when they are sounded simultaneously?
A)1%
B)2%
C)3%
D)4%
E)5%

Question

Quizplus · Accepted Answer

The beat frequency is given by the absolute difference in the frequencies of the two sources. Here, the beat frequency is 10 Hz, which means the frequency of one of the pipes must change to either 490 Hz or 510 Hz to achieve this difference. The fundamental frequency $f$ of a closed-open pipe is given by $f = \frac{v}{4L}$, where $v$ is the speed of sound and $L$ is the length of the pipe. Since the speed of sound $v$ remains constant, any change in frequency is inversely proportional to the change in length. To achieve a frequency of 510 Hz (a 2% increase from 500 Hz), the length of the pipe must decrease by 2% (since frequency and length are inversely related). Similarly, to achieve a frequency of 490 Hz (a 2% decrease), the length must increase by approximately 2%. Therefore, a 2% change in length is required to produce a 10 Hz beat frequency.

Two Identical Closed-Open Pipes Each Have a Fundamental Frequency of 500