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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.
 Fall '08
 SPRUNGER
 Physics, Energy, Kinetic Energy, Mass

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