P17_049.PDF

49. We consider an inﬁnitesimal segment of a string oscillating in a standing wave pattern. Its length is

d x and its mass is d m = µ d x ,where µ is its linear mass density. If it is moving with speed u its

kinetic energy is d K = 2 u 2 d m = 2 µu 2 d x . If the segment is located at x its displacement at time t is

y =2 y m sin( kx )cos( ωt ) and its velocity is u = ∂y/∂t =

2 ωy m sin( kx ) sin( ωt ), so its kinetic energy is

d K = 1

4 µω 2 y 2 m sin 2 ( kx ) sin 2 ( ωt )=2 µω 2 y 2 m sin 2 ( kx ) sin 2 ( ωt ) .

Here y m is the amplitude of each of the traveling waves that combine to form the standing wave. The

inﬁnitesimal segment has maximum kinetic energy when sin 2 ( ωt ) = 1and the maximum kinetic energy

is given by the diﬀerential amount

dK m =2 µω 2 y 2 m sin 2 ( 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

K m =2 µω 2 y 2 m L/n

sin 2 ( kx ) dx .

We use the trigonometric identity sin 2 ( kx )= 2 [1+ 2 cos(2 kx )] to obtain

K m = µω 2 y 2 m L/n

[1+ 2 cos(2 kx )] d x = µω 2 y 2 m L

n + 1

k sin 2 kL

For a standing wave of n loops the wavelength is λ =2 L/n and the angular wave number is k =2 π/λ =

nπ/L ,so2 kL/n =2 π and sin(2 kL/n ) = 0, no matter what the value of n .Thus,

K m = µω 2 y 2 m L

To obtain the expression given in the problem statement, we ﬁrst make the substitutions ω =2 πf and

L/n = λ/ 2, where f is the frequency and λ is the wavelength. This produces K m =2 π 2 µy 2 m f 2 λ .We

now substitute the wave speed v for fλ and obtain K m =2 π 2 µy 2 m fv .

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