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Unformatted text preview: CHAPTER 2
VIBRATION 0F STRINGS 21 W The vibration of strings has limited practical value but such a study
demonstrates a large variety of principles fundamental to wave propagation.
Consider the small transverse vibrations of a tightly stretched string in the x—y plane. Denote the transverse motion by y = y(x,t). The string is perfectly flexible‘ (no energy losses), and we assume that the tension T is a constant along the
string. Consider an element of string T
2?; d1!)
/ I
”I y
I!)
X
T
Applying Newton's Law in the y—direction gives
32
T sin(lp+dlb) — Tsimp = pds ~—§ , (2.1) at 22 where p is the mass/length of the string. (2.1) may be written as 2
T[sin(¢+d¢) — sinW] = pds §—§
8t
32
or T[sinw cosdw + cosw sindw — sinW] = pds—%§.
at
For small dw, the above becomes
82
T cosw dw = pds ——% . (2.2)
at
Now from geometry
3 2 32
tanw = 5X=%> sec w dw = —~% dx
X 3X
2 32
dw = cos w §~dx . (2.3)
EX Equations (2.2) and (2.3) give 2 2
dx T cost 3 X = p 8 X .
ds 2 2
3X 8t Assuming small deflections, g§~<< l, and SO,§§ 2 cosw and coszw = [l + (g%)2]_l E 1. So our final result becomes 2 2
9.2: $3.1
T 2“p 2
8X 3t
32 1 32
or —l>i=——Y, (2.4)
2 2 2
8X C St
where c = ‘2 .
9 Consequently the transverse vibrations of a string is governed by the one—
dimensional wave equation. Equation (2.4) together with initial and boundary conditions constitutes the boundary value problem for string vibrations. 22 WW RecallcnncD'Alembert solution to the one—dimensional wave equation
y(x,t) = f(X—ct) + g(x+ct) . (2.5) Consider a string of infinite extent with the following initial conditions y(X,0) F(X) (2.6) II §mm mm
Using (2.5) with (2.6) implies f(X) + g(X) = F(X) —cf'(x) + cg'(X) = V(X) ,
or integrating the second equation gives f(X) + g(X) = F(X) (
2.7) 4m+gm=ﬁﬁvmﬁ+k, where k is the constant of integration. Solving (2.7) for f and g gives %{F(x) — % [X V(€)d£ — k} g(X) = %{F(X) +% IX V(g)dg + k} f(X)
(2.8) Hence our solution is
l l X+ct
y(x,t) = E[F(x—ct) + F(x+ct)] + —2Cf v(g)dg . (2.9)
x—ct Consider the special example, I
{11 ll Y(X90) F(X) H
r—J‘ﬁ
Oi—‘O WA [I
O §mm ww 2—4 The solution for this case has the form y(x,t) =% [F(x—ct) + F(x+ct)] . (2.10) Graphically this solution at various instants of time looks like: 2—5 2.3 SEPARATION OF VARIABLES SOLUTION We now investigate the separation of variables solution for the iniiiaﬂ
dLApZacement phobﬁem of a string of finite length. The problem is formulated by the following boundary value problem. 1 .
yxx _ C2 ytt ’ O < X < k y(0,t) Y(%at) = O
(2.11) y(X,O) = f(X)
yt(x,0) = 0 . This problem represents a vibrating string fixed at X = O and X = 2 with an initial displacement but no initial velocity. Recall separating variables of (2.11)l gave y(X,t) = X(X) T(t)
X(X) = A coskx + B sinkx (2.12) T(t) Ccomﬂt + D sinckt. Using boundary conditions (2.11)2 implies that
X(O) = X(£) = O ,
which gives A = O B sinXQ = 0 2—6 Now B % 0, so sinkﬂ = O or A = X = **'. (2.13) An are normally called the eigenvaﬂueb of the problem. Boundary condition (2.11)4 yields that Hence for each n, we have a solution form y = sinREE cos cnﬂt n = 1, 2, 3, ... (2.14) n 2 2 These yn's are called the cigcnﬁunctionb of the problem. 1
E
1
l
1
1
i
1
1
i
r
T
1
l
S
E
1 Now the most general solution is found by a linear combination of forms (2.14), i.e., m . nﬁx nﬁct
y(X,t) =n§lbn Sln E*cos Q , (2.15) where bn are appropriate constants. The final boundary condition (2.11)3 then reads y(x,0) = f(x) =n:1 bn sin 2%E‘ . (2.16) f
E
But (2.16) is simply the Founien Aime éQﬂLeA tephcbcniation for f(x).
According to our theory the coefficients bn are then given by i 1 .Z 1  123; _
bn — Q f0f(€) Sin 2 d5 n ~ 1, 2, 3, ... (2.17) I So our solution to the given problem is (2.15) with the bn 8 given by (2.17). 2—7 24 FOURIER INTEGRAL SOLUTION Consider now a semi—infinite string stretched along the positive half of the x—axis with one end fixed at the origin. Assuming the string is released from rest With some prescribed initial displacement; the problem is formulated as _ _l__
yxx — C2 ytt ’ X~Z O
Y(O:t) = yt(x,0) = 0
(2.18)
Y(X,0) = f(X)
y(x,t) < M for some constant M (boundedness condition).
Separating the variables as before
Y(X,t) = X(X) T(t) ©
x" + AZX = o
(2.19)
x(0) = o , 1x(x) < Ml
.. 2 2
T + c A T — O (2.20) i<o>=o , 1T(t)<M2 , for some constants M1 and M2. Applying boundary condition (2.19)2 the solution of (2.19)l gives
X = sinKX Now sinkx is bounded for all X provided K is real, and since there are no other boundary conditions, A can take any real value. This means there is *
a canténuoub Aei of eigenvalues for this type of problem, Hence let
X = a, and our X solution is now (apart from an arbitrary constant multiplier)
X = sindx .
The solution to (2.20) is T = cosact The generalized linear combination of functions XT for all positive a (note negative a produces no new additional eigenfunctions) is y(x;t) = f: g(a) sinax cosact dd . (2.21)
Using the boundary condition (2.18)3 implies y(x,0) = f(x) = f: g(a) sinax dd . (2.22) But (2.22) is simply the Fourier sine integral representation for f(x). According to the theory of Fourier integrals the function g(a) is then given by
2 W .
gm) = E [O f(€) smagdg . (2.23)
Therefore our solution is y(x,t) =% f: sindx cosdct I: f(§) sindgdgdd . (2.24) 2.5 QQRRESPONDENCE BETWEEN SOLUTION TYPES We now will demonstrate the correspondence between the traveling wave solution and the separation of variables and Fourier integral solutions. * The separation of variables solution for the finite string produced a
diAcActe Aei of eigenvalues. 2—9 Separation of Variables Solution Consider again the problem _ _l_.
yxx — C2 ytt ’ O < X < 2
y(0,t) = yUht) = yt(x,0) = 0 Y(X:O) = f(X)
The traveling wave solution is given by (2.9) with V(X) = O, i.e.,
y(X,t) =§ [f(X—ct) + f(x+ct)] , (2.25) with f(x) = f(x+£).
Next assuming that f is an odd function, f(x) = f(—X)7expand f in its Fourier sine series representation _ l. w . nﬁ _ m . nﬁ
y(x,t) — 2{nElbn Sln [ﬂ (X ct)] +n£lbn Sln[£ (x+ct)]}
w nﬁx nﬁct
_nElbn sin 2 cos k , 2 £ . nﬂ
where bn = E'IO f(€) Sln —E§ d5 , which is the same as solution (2.15) Fourier Integral Solution Consider the problem .._1._
yxx _ C2 ytt ’ X: 0 Y(O,t) = yt(X’0) = 0 y(x,0) = f(X)
The traveling wave solution is Y(X,t) ='% [f(X—ct) + f(x+ct)] 2—10 Next assuming that f is an odd function,represent f in its Fourier sine integral representation y(x t) =‘l [w B(a) sina (x—ct)da +l fm B(a) sina (x+ct)da
’ 2 O 2 O I: B(a) sinux cosactda , Where B(d) = é‘f: f(€) sinagdg s which is the same as (2.24). 2.6 VIBRATION OF STRINGS WITH DAMPING Consider the string vibration case with external or internal damping
proportional to the velocity of the string. Summing forces on an element of string now gives 2 2 A §_X = l;.§_z.+.kn.§x. (2 26)
2 2 2 2 at ’ ' 8X C St c where h is the damping constant.
Equation (2.26) is known as the equation 06 Ickeghaphy. Let us try and separate the variables; i.e., let y(x,t) = X(X) T(t) . Relation (2.26 then implies that A
I! _ l_ " 3_ °
X T — 2 XT + 2 XT
c ' c.
X”__l :1: L]; — 2
or X — 2 [T + k T] — A .
c
Consequently, X" + AZX = 0 T + @i + AZCZT = 0 (2.27) The solutions to (2.27) are coskx
X = { sinAX} 2—11 T = exp E% (—k i V £2 ~ 4X2c2)t] Note that for most cases/122 * 4l2c2 < O (underdamped case) and so the
T—solution would reduce to
coth
T={ }ekt/2,
sinBt
A
where 82 = Azcz — %~k2. So our general solution to (2.26) is of the form coslx}{coth}e~Qt/2 sinkx sinBt (2'28) Y(X9t) = { 27 VIBRATION OF A COMPOSITE STRING We now want to solve the problem of the transverse vibrations of two semiinfinite strings of different densities joined together at the origin. Denote the displacement of the two strings by yl and y2. Suppose that
a train of harmonic waves is incident from the negative x—direction. When
these waves meet the change in string density, they will suffer pamtéaﬁ deﬂection and lama/(7, MammX/szswn. Hence II +
yl yincident yreflected (2.29) ll y2 ytransmitted . Let us represent all wave motion in the compﬂcx cxponcniiaﬁ ﬁonm, i.e. I
1
l
J
I
I 2—12 _ 21Ti(mt—k1X)
yincident ‘ Ale B e2wi(wt+klx) yreflected = 1 (230) _ 2wi(wt—k2x)
ytransmitted _ AZe The amplitude A is real, but A and B1 may be complex. All three waves 1 2 must have the same frequency w, but since the velocities in the two strings
are different, the transmitted wave will have a different wavelength than the incident and reflected waves. The wave velocities are given by c = i} , c = (2.31) 9_
l l 2 k2 Note since c2 ='§ , we find that NIH
Ill (2.32) In order to determine A2 and B1 we use the matching or continuity conditionb at x = 0, i.e., y1(0,t) = y2(0,t)
3yl(0,t) 8y2(0,t) (2.33) 8x = ESE. Using (2.29) and (2.30) in (2.33) implies that A1 + B1 = A2 2wi(—klA (2.34) l + lel) = 2Wi(—k2A2) Solving (2.34) for A and B1 yields 2 2kl
A. =——————A
2 kl + k2 1 B =E:1_*EigA
l kl + k2 l (2.35) 2—13 Since Al’ kl and k2 are real, so are A2 and B1. Note A2 is always positive, but B1 is p051tive only if kl > k2. Define the coeMLcéen/t 05 deﬂection, R = W incident energy El 2 ksz 2 51‘— /p.2— 2
R = A— = (k +k) = (m) , (2.36)
1 1 2 E + V5 transmitted energy and the coeﬁﬁicéen/t 06 Wmmbion, T = , .
1nc1dent energy ”0—1/8 T=1—R_————————2
(EM/52‘) (2.37) M W The ﬁlow of energy is associated with wave motion. Consider the Vibrating string situation. The kinetic enengy of vibration is given by K = f—i—m'rzdx . (2.38) 2
The poicniiaﬁ enengg V is found by considering the increase in length of an element of string. deformed string element / X original length The element has increased its length from dx to ds. Hence the work done
on this element is T(ds—dx), where T is the tension. Summing for all elements of the string gives f T(ds—dx) = g T(de2 + dy2 l dx)
2
{TG/l + (13:1)2 — l)dx = —§ T f (@1)zdx, (2.39) 2 <
II II 2—14 for small slopes where 3% << 1. For the case of p&OQ&QAALve Oh taaveking waucb y = f(x i ct), (2.38) and (2.39) become 2 K = % pc (f')2dx = %~T f (f')2dx
x NIH 29‘. T f (f')2dx . (2.40)
2 Thus the kinetic and potential energies are equal. The above result does
not apply however to stationary waves or any combination of progressive waves traveling in opposite directions. For the special case of a hanmonic thaveﬁing wave y = ¢ = A cos(kX ~ wt) , (2.41)
we have that dK = dV =% Azpwzsin2(kx — wt) , (2.42)
and so the total energy per unit length is U = dK + dV = Azpwzsin2(kx — wt) . (2.43) For a Mariana/Ly wave y = ¢ = Asinkxsin(wt + 8) , (2.44)
we have dK = éAzpwzsinzkxcosz(wt + 8) dV = %—A2pw2c082kxsin2(wt + a) (2.45)
and dK % dV. From the above discussion it should be clear that for any type of wave
motion the total energy density at any point varies with time. This fact implies the existence of a ﬁﬁux, carrying energy into and out of an arbitrary 2—15 element. The change in energy between times t and t + dt of an element of length dx at position X will be given by [U(x,t + dt) — U(X,t)]dx = 3311 dxdt , (2.46) where U is the energy/length for the string problem. This will be equal to the net energy that has been carried into the element in the time dt, [P(X,t)  P(x+dx,t)]dt = — 3? dxdt , (2.47) where P is the powcn of the energy flux, i.e., energy per unit time. Applying conservation of energy 3‘1 = — ~33 . (2.48) For the case of traveling waves, U = U(xict) and P = P(Xict), so (2.48) reduces to P' = ; cU' . Hence P = + cU , (2.49) where the sign gives the direction of energy flow. Consequently the energy that passes any point in a time interval dt is the energy contained in an element of length cdt. E
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