A semi-infnite string is driven by a orce at a distance L rom L rom the the fxed fxed end. What What is the the mech mechani anica call impe impeda danc nce e seen seen by the orce? orce? Interpret the individual terms in your answer. Solution y "t!
y #t!
f t!
We will divide the string into two regions: the frst lies between the wall and the applied orce$ and the second stretches rom the applied orce out to infnity infnity. %ince the applied applied orce causes causes a discontinuity discontinuity in the string$ we must must defne defne two separate separate uncti unctions: ons: y " hold holds s over over the frst regio region n and and y # holds over the the second region. region. We begin by using using the solution we ound ound in class or the string j ( ωt −kx )
+ B e j (ωt +kx)
j ( ωt −kx )
+ D e j( ωt +kx)
y 1 ( x , t )= A e y 2 ( x , t )=C e
&ut the &ut the righ rightt part part o the the stri string ng exte extend nds s to infni infnity ty$$ so that that no wave wave is re'ected in the (x direction over this part o the string) that is$ D * +. We have have thre three e un,n un,now own n coe coeic icie ient nts$ s$ so we must must defn defne e thre three e boun bounda dary ry conditions. he frst is the fxed wall y 1 ( 0, t )=0
he second is defned by the act that the two solutions must e/ual each other at the applied orce
y 1 ( L , t ) = y 2 ( L , t )
he fnal boundary condition is defned by drawing a ree-body diagram at the point o application o the orce y #t!
y "t!
T
T f t!
%umming orces at this point gives ∂ y1 ∂y f ( t )−T + T 2 =0 ∂ x L ∂ x L
|
|
%ubstituting the frst boundary condition into y" gives B =− A
hus y 1 ( x ,t )=( e
− jkx
−e jkx ) e jωt
¿− 2 j A sin kx e jωt %ubstituting the second boundary condition gives
−2 j A sin kL e jωt =C e− jkL e jωt A =
jC e
− jkL
2sin kL
And fnally$ applying the third boundary condition gives
(
F + T −2 jk ∙
jC e
− jkL
2sin kL
cos kL
− jkL
F −kT C cot kL e
)−
jkT C e
− jkL
− jkT C e− jkL =0
=0
− jkL
F −kT C ( cot kL + j ) e C gives
%olving or C =
=0
F − jkL
kT ( cot kL + j ) e
kL cos kL− j sin ¿
¿ ( kT cot kL + j ) ¿ C =
F
¿
kL cos kL− j sin ¿ kT
(
¿
sin kL
C =
C =
+ j ¿
F
¿ F
(
kT
C =
)
cos kL
2
cos kL sin kL
− j cos kL + j cos kL + j sin kL
)
F sin kL kT ( cos kL+ sin kL ) 2
2
F sin kL C = kT
although not necessary$ we can use this result to solve or A as well − jkL
je
A =
F sin kL kT 2sin kL
A =
j F 2 kT ( cos kL + j sin kL )
∙
he velocity at the point o application o the orce is u ( L, t )= jω C e
j ( ωt − kL )
− jkL
jω F sin kL e u ( L, t )= kT
e
jωt
hus$ the impedance is Z =
Z =
Fe
jωt − jkL
jω F sin kL e kT
jωt
e
kT jω sin kL e
− jkL
Z =
kT ( cos kL + j sin kL ) jω sin kL
Z =
kT ( cot kL + j ) jω
Z =
−kT
ω
cot kL +
kT ω
Z =− j ρ L c cot kL + ρ L c
he frst term is the impedance o a string with one fxed end while the second term is the impedance o a semi-infnite string. 0learly$ impedances in parallel are additive.
F
m
s
b
Problem 2:
A semi-infnite string is attached to a mass1spring1damper system. What is the mechanical impedance seen by a orce driving the mass? Interpret the individual terms in your answer. Solution f t! y 0,t! m
f c
T
f k
2irst$ draw a ree-body diagram o the mass. he string is semi-infnite$ so its displacement unction is y ( x , t ) = A e
j( ωt − kx )
he slope at x * + is
|
∂y =− jk A e jωt ∂x 0
%umming orces on the mass gives m ´ y ( 0, t ) =−c ´ y ( 0, t )− ky ( 0, t )+ T
he time derivatives are
|
∂y + f (t ) ∂x 0
y´ ( 0, t )= jω A e
jωt
y´ ( 0, t )=−ω A e 2
jωt
%ubstituting these into the e/uation o motion gives
−mω 2 A e jωt + jωc A e jωt + k A e jωt + Tjk Ae jωt = F e jωt
−mω 2 A + jωc A + k A + jkT A = F
%olving or A gives A =
F
−mω 2+ jωc + k + jkT
he driving point velocity is y´ ( 0, t )=
jωFe
jωt
2
−mω + jωc + k + jkT
hus$ the driving point impedance is jωt
Fe Z = jωt jω F e −mω 2+ jωc + k + jkT 2 m ω + jωc + k + jkT − Z =
jω
Z =
(
)
jm 2 jωc ω− −ω 2n + ρ0 c ω m
he frst term is the impedance o the mass1spring1damper system while the second term is the impedance o the semi-infnite string. As beore$ impedances in parallel are additive.
