Question

You are playing a note that has a fundamental frequency of \(400 .\) Hz on a guitar string of length \(50.0 \mathrm{~cm} .\) At the same time, your friend plays a fundamental note on an open organ pipe, and 4 beats per second are heard. The mass per unit length of the string is \(2.00 \mathrm{~g} / \mathrm{m} .\) Assume that the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). a) What are the possible frequencies of the open organ pipe? b) When the guitar string is tightened, the beat frequency decreases. Find the original tension in the string. c) What is the length of the organ pipe?

Short answer

Answer: The possible frequencies of the open organ pipe are 396 Hz and 404 Hz. The original tension in the string is 32 N, and the length of the organ pipe is approximately 0.216 m.

Step-by-step solution

Find the beat frequencies

Given the beat frequency as 4 Hz, we can write the two possible scenarios for the organ pipe's frequency: \(f_{organ1} = f_{guitar} - 4 = 400 - 4 = 396\) Hz \(f_{organ2} = f_{guitar} + 4 = 400 + 4 = 404\) Hz We now have the possible frequencies for the organ pipe. #b) Original tension in the string:#

Determine the frequency of the string based on beat frequency change

We are told that when the guitar string is tightened, the beat frequency decreases. Hence, the original frequency of the open organ pipe is the one with a lower frequency than the guitar note: \(f_{organ} = 396\) Hz

Find the tension in the string

Using the formula for frequency of a vibrating string, we get: \(400 = \frac {1} {2(0.5)} \sqrt{\frac {T} {0.002}}\) Solve for \(T\): \(T = 0.002\left(\frac {2(0.5)(400)} {1}\right)^2\) \(T = 32\) N So, the original tension in the string is 32 N. #c) Length of the organ pipe:#

Find the length of the organ pipe

Use the formula for the frequency of an open organ pipe with \(f_{organ} = 396\) Hz: \(396 = \frac {1} {4L} (343)\) Solve for \(L\): \(L = \frac {1} {4 (396)} (343)\) \(L = 0.216\) m The length of the organ pipe is approximately 0.216 m.