Two identical closed-open pipes each have a fundamental frequency of 500 Hz.If the length of one of the pipes is increased by 1% What is the beat frequency when they are sounded simultaneously?
A)2 Hz
B)8 Hz
C)10 Hz
D)13 Hz
E)5 Hz

Question

Quizplus · Accepted Answer

The fundamental frequency $f$ of a closed-open pipe is inversely proportional to its length $L$, i.e., $f \propto \frac{1}{L}$. If the length of one pipe is increased by 1%, its new length becomes $1.01L$, and its new frequency $f'$ becomes $f' \propto \frac{1}{1.01L}$. This means the new frequency is $f' = \frac{f}{1.01}$. Given $f = 500$ Hz, $f' = \frac{500}{1.01} \approx 495.05$ Hz. The beat frequency is the absolute difference between the two frequencies, which is $500 - 495.05 = 4.95$ Hz, which rounds to approximately 5 Hz.

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