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A string with 2 fixed ends. The third harmonic is observed to be at a frequency...

A string with 2 fixed ends. The third harmonic is observed to be at a frequency of 120Hz. The string has a mass/length of 0.4 kg/m and is under a tension of 40 N.

1. What is the length of the string?

2. What is the fundamental frequency and wavelength for this oscillator?

3. Find the wavelengths and frequencies for the 2nd, 4th, and 5th harmonics. How far from the end of the string is the first anti-node of the 5th harmonic?

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Answer #1

Wave speed is given by Here T is the tension and u is the linear mass density Now speed of the wave is T 40 = 10 x m/s 0.4 AL 2L 2 = 2. = 7 SO V 2L 3 3v L= (3) (10) L = (2) x (120) L = 0.125m 2. In fundamental mode */ = 1 2 = 2L fundamental frequencdhi dhe 2 = 1 wavelength for 2nd harmonic 2 = 0.125m frequency for 2nd harmonic f- f= 10 (0.125) f = 80Hz 4th harmonic din dh0.125 2 = = 0.0625m frequency for 4th harmonic 2v f= ܢܠ L f 2 x (10) (0.125) f = 160Hz 5th harmonic Niz al d/ dla d/2 + + + Ld/ dla Al2 t + + + L First anti-node of the 5th harmonic is at a distance of d = 4 From the end d = 0.05 4 = 0.0125m First

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