A stretched string is fixed at both ends which are 600 cm apart. If the density...
A stretched string is fixed at both ends which are 600 cm apart. If the density of the string is 0.028 g/cm, and its tension is 600 N, what is the wavelength of the 5th harmonic? Answer in units of cm.
Homework Answers
For a string (or a tube open at both ends),
λ_n = 2L / n → for n a positive integer. n = 1 is the "fundamental."
For the 5th harmonic
λ_5 = 2 * 600cm / 5 = 240 cm ◄
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A stretched string is fixed at both ends which are 600 cm apart.
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