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 seconds are heard. The mass per unit length of the string is \(2.00 \mathrm{~g} / \mathrm{m}\). Assume the velocity 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?

To find the possible frequencies of the open organ pipe, we need to consider the beat frequency that we hear. The beat frequency is the absolute difference between the frequencies of the two sources. It is given as 4 beats per second. If f1 is the frequency of the guitar string and f2 is the frequency of the open organ pipe, we can represent the beat frequency as: abs(f1 - f2) = 4 Hz We are given f1 (fundamental frequency of the guitar string) as 400 Hz. Let's solve for the possible values of f2. f2 could either be 4 Hz greater or 4Hz less than the guitar string's fundamental frequency. 1. f1 + 4 = f2 400 + 4 = 404 Hz 2. f1 - 4 = f2 400 - 4 = 396 Hz So, the possible frequencies of the open organ pipe are 404 Hz and 396 Hz.

To find the original tension in the guitar string, we need to use the formula for the fundamental frequency of a vibrating string. This formula is given by: f1 = (1/2L) * sqrt(T/μ) Where f1 is the fundamental frequency (400 Hz), L is the length of the string (0.5 m), T is the tension, and μ is the mass per unit length (0.002 kg/m). Now, we need to rearrange the formula to solve for T and plug in the given values: T = (2Lf1)^2 * μ T = (2 * 0.5 * 400)^2 * 0.002 T = (400)^2 * 0.002 T = 320 N The original tension in the string is 320 N.

To find the length of the organ pipe, we first need to know the fundamental frequency of the organ pipe. Since tightening the guitar string decreased the beat frequency, we should choose the smaller frequency for the organ pipe, which is 396 Hz. The formula for the fundamental frequency of an open organ pipe is: f2 = v / (2L_pipe) Where f2 is the fundamental frequency of the open organ pipe (396 Hz), v is the velocity of sound (343 m/s), and L_pipe is the length of the organ pipe. Now let's rearrange the formula to solve for L_pipe and plug in the given values: L_pipe = v / (2f2) L_pipe = 343 / (2 * 396) L_pipe ≈ 0.432 m The length of the organ pipe is approximately 0.432 m or 43.2 cm.