An air column in a tube of length L = 50 cm, | vedclass

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(C) For a tube closed at one end,the frequencies of harmonics are given by $f_n = n \frac{v}{4L}$,where $n = 1, 3, 5, ...$ are odd integers. The fifth

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(C) For a tube closed at one end,the frequencies of harmonics are given by $f_n = n \frac{v}{4L}$,where $n = 1, 3, 5, ...$ are odd integers. The fifth harmonic corresponds to $n = 5$.The wavelength $\lambda$ is given by $\lambda = \frac{4L}{n} = \frac{4 	imes 50 \ cm}{5} = 40 \ cm$.The standing wave equation for a closed pipe (taking the open end at $x = 0$) is $y = A \sin(kx) \cos(\omega t)$,where $k = \frac{2\pi}{\lambda} = \frac{2\pi}{40} = \frac{\pi}{20} \ rad/cm$.The particle at the open end $(x = 0)$ is at a node of the displacement wave (or antinode of pressure),but in terms of displacement amplitude,the open end is an antinode. However,the standard equation $y = A \sin(kx) \cos(\omega t)$ assumes $x=0$ is a node. For an open end at $x=0$,the displacement is $y = A \cos(kx) \cos(\omega t)$.At $x = 0$,$y_1 = A \cos(0) \cos(\omega t) = A \cos(\omega t)$.At $x = 42 \ cm$,$y_2 = A \cos(k 	imes 42) \cos(\omega t) = A \cos(\frac{\pi}{20} 	imes 42) \cos(\omega t) = A \cos(2.1\pi) \cos(\omega t) = A \cos(2\pi + 0.1\pi) \cos(\omega t) = A \cos(0.1\pi) \cos(\omega t)$.Since both particles oscillate in phase (the cosine terms have the same sign),the phase difference is $0^{\circ}$.

An air column in a tube of length $L = 50 \ cm$,closed at one end,is vibrating in its fifth harmonic. The phase difference between a particle at the open end and a particle at $42 \ cm$ from the open end is: (in $^{\circ}$)

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