Question

Resonance Pre-Lab Assignment (1 point) Recall from the "Introduction to Waves" lab that it was easy to calculate the harmonic number (n) and wavelength (A) of standing waves on a string by counting the number of antinodes (n -21/n). That was a system with nodes fixed at the end points. Today you will be working with a system that has one open end and one closed end (i.e. a node fixed at one end and an antinode fixed at the other end). You can still figure out the harmonic number and wavelength of standing waves, but it is different from before. In this system: 2n - 4L/n. Discuss why it is only possible to produce the odd harmonics in a system with one open end and one closed end (pictures will helpl). Write down the relation between h and L for the n 1, 3, 5, and 7 harmonics, and calculate the wavelength and frequency of the fundamental resonant mode for L # 0.875 m and b 686 Hz. Then calculate the wave velocity of this oscillation. Pipe with one open and one closed end " Only the odd harmonics can be produced ts Resonant Mode to open end closed end Fundamental 2nd Harmonic3 to in 3) 3rd Harmonic5 fo 4th Harmonic 7 fo Length L PHYS 2LC: Lab 3
I need help with the 3,5,7 harmonics part of the question

Answer 1

Since closed end is constrained to be a node of the wave and the open end is of antinode hence only odd harmonic is possible For n = 1: lambda_1 = 4 L = 3.5 m n = 3: 3 lambda /4 = L = > lambda_3 = 4 L/3 = 1.16 m n = 5: 5/4 lambda = L = > lambda_5 = 4 L/5 = 0.7 m n = 7: 7/4 lambda = L = > lambda_7 = 4 L/7 = 0.5 m and we know that = n v/4 L for n = 1 f_7 = 7 v/4 L = 686 = > V/4 L = 686/7 V/4 L = 98 Hz Hence f_1 = 98 Hz: f_3 = 3 f_1 = 294 Hz f_5 = 490 Hz \r\n Since we know that V_1 = f_1 lambda_1 = 98 times 3.5 = 343 m/s V_3 = f_3 lambda_3 = 294 times 1.167 = 343 m/s V_5