P17_049 - 49. We consider an infinitesimal segment of a...

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Unformatted text preview: 49. We consider an infinitesimal segment of a string oscillating in a standing wave pattern. Its length is dx and its mass is dm = µ dx, where µ is its linear mass density. If it is moving with speed u its kinetic energy is dK = 1 u2 dm = 1 µu2 dx. If the segment is located at x its displacement at time t is 2 2 y = 2ym sin(kx) cos(ωt) and its velocity is u = ∂y/∂t = −2ωym sin(kx) sin(ωt), so its kinetic energy is dK = 1 2 2 2 4µω 2 ym sin2 (kx) sin2 (ωt) = 2µω 2 ym sin2 (kx) sin2 (ωt) . Here ym is the amplitude of each of the traveling waves that combine to form the standing wave. The infinitesimal segment has maximum kinetic energy when sin2 (ωt) = 1 and the maximum kinetic energy is given by the differential amount 2 dKm = 2µω 2 ym sin2 (kx) . Note that every portion of the string has its maximum kinetic energy at the same time although the values of these maxima are different for different parts of the string. If the string is oscillating with n loops, the length of string in any one loop is L/n and the kinetic energy of the loop is given by the integral L/n sin2 (kx) dx . 2 Km = 2µω 2 ym 0 2 We use the trigonometric identity sin (kx) = 1 2 [1 + 2 cos(2kx)] to obtain L/n 2 Km = µω 2 ym 2 [1 + 2 cos(2kx)] dx = µω 2 ym 0 2kL L1 + sin n k n . For a standing wave of n loops the wavelength is λ = 2L/n and the angular wave number is k = 2π/λ = nπ/L, so 2kL/n = 2π and sin(2kL/n) = 0, no matter what the value of n. Thus, Km = 2 µω 2 ym L . n To obtain the expression given in the problem statement, we first make the substitutions ω = 2πf and 2 L/n = λ/2, where f is the frequency and λ is the wavelength. This produces Km = 2π 2 µym f 2 λ. We 2 2 now substitute the wave speed v for f λ and obtain Km = 2π µym f v . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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