PHYS 422 Lecture 21 Continuous Systems and Fourier Analysis I

# PHYS 422 Lecture 21 Continuous Systems and Fourier Analysis I

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t i t d Continuous Systems and ourier Analysis Fourier Analysis I

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Continuous Systems a stretched string of length L assumptions: the mass per unit length is a constant 0L and the string is elastic ravitational forces are negligible M L μ= gravitational forces are negligible there are only small deflections in the transverse direction the mass of the element above is μΔ x since there are no horizonital displacements T T in the transverse direction, the net force is 11 2 2 cos cos TT θθ = = 2 1 1 sin sin θ Newton II 22 2 1 1 2 sin sin y x t μ −= Δ
Continuous Systems utilize the horizontal tension components 22 1 1 21 sin sin tan tan s cos TT T θ −= 1 1 2 2 cos tan tan xy t θθ μ Δ∂ Tt tan θ 1 and tan θ 2 are the slopes of the string at x and x+ Δ x y ⎛⎞ bstitute and divide by 12 tan tan x xx yy ∂∂ = = ⎜⎟ ⎝⎠ substitute and divide by Δ x 2 2 1 y x x T t ⎡⎤ ⎢⎥

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## This note was uploaded on 03/23/2011 for the course PHYS 422 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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PHYS 422 Lecture 21 Continuous Systems and Fourier Analysis I

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