3- 37 Problems and Solutions Section 3.3 (problems 3.26-3.32) 3.26 Derive equations (3.24). (3.25) and (3.
26) and hence verify the equations for the Fourier coefficient given by equations (3.21), (3.22) and (3.23). Solution: For n ! m, integration yields:
T
!
0
$ sin n # m " T t sin n + m " T t ' sin n" T t sin m" T tdt = & # ) " 2 n # m " 2 n + m & ) T % T (0
) ( ) ' $ ' sin $ %( n # m) ( 2* ) ( # sin %( n + m) ( 2* ) ( = 0 = 2 ( n # m)" 2 ( n + m)"
T T
$ $ + 2* . ' + 2* . ' sin & n # m - 0 T ) sin & n + m - 0 T ) , T / ( , T / ( % % = # 2 n # m "T 2 n + m "T
Since m and n are integers, the sine terms are 0, so this is equal to 0. Equation (3.24), for m = n:
T
!
0
$1 ' $ + 2* . ' 1 T T sin n" T tdt = & t # sin 2 n" T t ) = # sin & 2* - 0 T ) 4 n" T % , T / ( %2 ( 0 2 8n*
2
T T T # sin $ 4 n* ' = % ( 2 8n* 2
Since n is an integer, the sine term is 0, so this is equal to T/2.
T
So,
!
0
$ 0 m# n sin n" T t sin m" T tdt = % &T / 2 m = n
Equation (3.25), for m ! n
3- 38
% sin n # m " T t sin n + m " T t ( cos n" T t cos m" T tdt = ' # * 2 n # m " 2 $ + m " ' * T T & )0
!
0
( (
) )
( (
) )
% % , 2+ / ( , 2+ / ( sin ' n # m . 1 T * sin ' n + m . 1 T * - T 0 ) - T 0 ) & & = # 2 n # m "T 2 n + m "T
( ) ( ) ( % ( sin % &( n # m) ( 2+ ) ) # sin &( n + m) ( 2+ ) ) = 0 = 2 ( n # m)" 2 ( n + m)"
T T
Since m and n are integers, the sine terms are 0, so this is equal to 0. Equation (3.25), for m = n becomes:
#1 & # * 2) - & 1 T T cos n " tdt = t + sin 2 n " t = + sin % ( % 2n , / T ( T T ! 2 4 n " 2 8 n ) $ + T . ' 0 $ '0 T
T 2
T T T + sin # 4 n) & = $ ' 2 8n) 2
Since n is an integer, the sine term is 0, so this is equal to T/2.
T
So,
!
0
$ 0 m# n cos n" T t cos m" T tdt = % &T / 2 m = n
Equation (3.26), for m ! n :
T
!
0
$ cos n # m " T t cos n + m " T t ' cos n" T t sin m" T tdt = & # ) 2 " n # m 2 " n + m & ) T T % (0
$ $ + 2* . ' + 2* . ' cos & n # m - 0 T ) cos & n + m - 0 T ) , T / ( , T / ( 1 1 % % = # # + 2 n # m "T 2 n + m "T 2 m # n "T 2 m + n "T
' $ ' cos $ 1 1 % n # m 2* ( # cos % n + m 2* ( # + =0 2 n # m "T 2 n + m "T 2 m # n "T 2 m + n "T
( (
)( ) )
( (
)( ) )
Since n is an integer, the cosine term is 1, so this is equal to 0.
3- 39
T
So,
!
0
cos n" T t sin m" T tdt = 0
# 1 & T cos n" T t sin n" T tdt = % sin 2 n" T t ( = sin 2 2) n = 0 $ 2 n" T '0 4 n)
T
T
Equation (3.26) for n = m becomes:
T
!
0
Thus
!
0
cos n" T t sin n" T tdt = 0
3- 40 3.27 Calculate bn from Example 3.3.1 and show that bn = 0, n = 1,2,,! for the triangular force of Figure 3.12. Also verify the expression an by completing the integration indicated. (Hint: Change the variable of integration from t to x = 2"nt/T.) Solution: From Equation (3.23), bn =
2 T
! F (t ) sin n"
0
tdt . Computing the integral yields:
2* bn = , T, + bn =
T /2
!
0
#4 & t " 1( sin n) T tdt + % $T '
T /2
* 4 # T&,1 " % t " ( / sin n) T tdt / 2'. / + T$ .
T T 4 t sin n) T tdt / ! T T /2 / .
2 *4 , T, +T
T /2
!
0
T /2
t sin n) T tdt "
!
0
sin n) T tdt + 3 ! sin n) T tdt "
T /2
Substitute x = n! T t =
2" n t T
!n
1 $ 2 bn = & !n & %! n =
!n
"
0
x sin xdx #
"
0
sin xdx + 3 "
!n !n
2! n
!n
2 sin xdx # !n
2! n
2! n
!n
"
' x sin xdx ) ) (
' ) 0 !n ( ' 1 $ 2 2 = #! n cos ! n + cos ! n # 1 # 3 + 3cos ! n # #2! n + ! n cos ! n ) & ! n %! n !n ( 1 1 $ $0 ' = 0 = #2cos ! n + 4cos ! n # 4 + 4 # 2cos ! n ' = % ( !n !n % (
1 $ 2 sin x # x cos x !n & %! n
+ cos x 0 # 3cos x ! n #
2 sin x # x cos x !n
2! n
From equation (3.22), an =
2 T
! F (t ) cos n"
0
T
tdt
2* an = , T, +
T /2
!
0
#4 & t " 1 cos n) T tdt + % ( $T '
T /2
* 4 # T&,1 " % t " ( / cos n) T tdt / 2'. / + T$ .
T T 4 cos n) T tdt " ! t cos n) T tdt / T T /2 / .
2 *4 an = , T, +T
T /2
!
0
T /2
t cos n) T tdt "
!
0
cos n) T tdt + 3 !
T /2
Substitute x = n! T t =
2" n t T
3- 41
an = = = =
1 $ 2 & !n & %! n
1 $ 2 !n & %! n 1 $ 2 !n & %! n
2
( (
' x cos xdx ) ) 0 0 !n !n ( 2! n ' !n 2! n 2 cos x + x sin x # sin x 0 + 3sin x ! n # cos x # sin x ) !n !n ( ' 2 cos ! n # 1 # 1 # cos ! n ) !n (
!n
"
x cos xdx #
!n
"
cos xdx + 3 "
2! n
cos xdx #
2 !n
2! n
"
2 $cos ! n # 1 # 1 + cos ! n ' ( ! n2 % * 0 4 , = 2 2$ cos ! n # 1' = + -8 % ( ! n , 2 2 -! n n even n odd
3- 42 3.28 Determine the Fourier series for the rectangular wave illustrated in Figure P3.28.
Solution: The square wave of period T is described by $1 0!t !" F t =%  " ! t ! 2"
()
Determine the coefficients a0 , an , bn from direct integration:
2 a0 = T
T
! F (t ) dt
0
an =
2 T
! ()
0
" 2" ' 2 $ = & ! 1 dt + ! #1 dt ) 2" & ) " %0 ( 2" 1 " = $t 0 # t " dt ' & ) ( "% 1 1 = $ " # 2" + " ' = 0 * a0 = 0 % ( " " 2# 2# F t cos n" T tdt , where " T = = =1 T 2#
()
( ) ()
If n is even, cosn" = 1. If n is odd, cosn" = -1 " 0 n even $ So, bn = # 4 $ ! n n odd % Thus the Fourier Series collapses to a sine series of the form
# 2# ( 1 %1 # 2# ( 2 % 1 = ' ! cos ntdt $ ! cos ntdt * = ' sin nt 0 $ sin nt # * 2# ' n * ) # &0 ) # &n 1 %sin n# $ sin n2# + sin n# ( = 0 = ) #n & T # 2# ( 2 2 % bn = ! F t sin " T tdt = ' ! sin ntdt $ ! sin ntdt * T 0 2# ' * # &0 ) # 2# ( 1 % $1 1 1 2 % %1 $ cos n# ( = ' cos nt 0 $ cos nt # * = $ cos n# + 1 $ 1 $ cos n# ( = & ) ) #&n n #n & ) #n
( )
( )
()
3- 43
F t = " bn sin nt =
n =1
()
4 sin nt n = 1,3,! n#
"
The Vibration Toolbox can also be used: t=0:pi/100:2*pi-pi/100; f=-2*floor(t/pi)+1; vtb3_3(f',t',100) [a,b]=vtb3_3(f',t',100) Note that vtb3_3 always gives some error on the order of delta t (.01 in this case). Using a smaller delta t reduced the error.
3- 44 3.29 Determine the Fourier series representation of the sawtooth curve illustrated in Figure P3.29.
Solution: The sawtooth curve of period T is 1 F t = t 0 " t " 2! 2! Determine coefficients a0 , an , bn :
()
2 a0 = T =
2 an = T
T
! ()
0
2 F t dt = 2"
2"
!
0
# 1 & # 1 &1 t ( dt = % 2 ( t 2 % $ 2" ' $ 2" ' 2
2"
1 * 4" 2 ) 0 , = 1 2 + 4"
tdt , where " T = 2# 2# = =1 T 2#
! F (t ) cos n"
0
2 = 2#
2# * 2# $ 1 ' 1 * t ) cos ntdt / = t cos ntdt ,! & , / ! 2 , / / + 0 % 2# ( . 2# , +0 . 2#
1 *1 1 1 *1 1 = cos nt + t sin nt / = 10 1 + 0 0 0 / = 0 2 , 2 2 , 2 n n 2# + n 2# + n .0 .
2 bn = T
! ()
0
2# 2# 2 * $ 1 ' 1 * F t sin n" T tdt = t ) sin ntdt / = t sin ntdt ,! & , / ! 2 2# , / / + 0 % 2# ( . 2# , +0 . 2#
1 = 2# 2 =
*1 1 1 , 2 sin nt 0 t cos nt / = n 2# 2 +n .0
*1 1 , 2 0 0 0 0 2# 0 0 / n +n .
1 $ 02# ' 01 = % n ) ( #n 2# 2 &
Fourier Series
F t = F t =
() ()
1 ! % #1 ( +" sin nt 2 n =1 ' & $n* ) 1 1 ! 1 # " sin nt 2 $ n =1 n
3- 45 3.30 Calculate and plot the response of the base excitation problem with base motion specified by the velocity
! t = 3e! t / 2 "(t ) m/s y
where #(t) is the unit step function and m = 10 kg, $ = 0.01, and k = 1000 N/m. Assume that the initial conditions are both zero. Solution: Given:
! t = 3e! t / 2 t m/s y m = 10 kg, " = 0.01, k = 1000 N/m ! 0 =0 x 0 =x
()
()
()
() ()
From Equation (2.61):
!! + c x !! y ! +k x! y =0 mx !! + cx !! + kx = cy ! + ky mx Integrate by parts to find y(t): ! t dt = 3e" t / 2 t dt y t =! y
) (
()
()
()
Let
() du = " ( t ) dt
u= t
dv = 3e! t / 2 dt v = !6e! t / 2
When
t > 0, t = 1, so y t = 6 1 ! e!1/ 2 !! + cx ! + kx = c 3e! t / 2 + 6 k 1 ! et / 2 So, mx
Since c = 2! km = 2 kg/s,
()
() (
!! + 2 x ! + 1000 x = 6000 ! 5994e! t / 2 10 x
The solution is given by equation (3.13):
3- 46
x t =
()
1 "#! n t & #! % e F % e n sin ! d t " % ( $ ' ) d% m! d 0 k = 10 rad/s m
()
!n =
! d = ! n 1 " # 2 = 10 rad/s
F t = 6000 " 5994e" t / 2 x t =
()
() ()
1 "0.1t & e $ 6000 " 5994e"% / 2 e0.1% sin 10 t " % ( d% ' ) 100 0
"0.1t t t * . , 0.1t ( d% " $ e"0.4t sin &10 t " % ( d% , 10 t " % + $ e sin & ' ) ' ) / ,0 , 0 0
( (
))
x t = 60e
After integrating and rearranging
x t = 6 ! 5.979e! t / 2 ! 0.0295cos10t ! 0.2990sin10t m
()
3- 47 3.31 Calculate and plot the total response of the spring-mass-damper system of Figure 2.1 with m = 100 kg, $ = 0.1 and k = 1000 N/m to the signal of Figure 3.12, with maximum force of 1 N. Assume that the initial conditions are zero and let T = 2" s.
Solution: Given: m = 100 kg, k = 1000 N/m,! = 0.1,T = 2" s, Fmax = 1N ,
! 0 = 0, # n = x 0 =x
() ()
k = 3.16 rad/s, # d = # 1 $ ! 2 = 3.15 rad/s, m From example 3.3.1 and Figure 3.10, $ 0 n even ! & F t " an cos nt , an = % -8 n =1 & 2 2 n odd '# n
#T =
2" = 1 rad/s T
()
!! + cx ! + kx = " an cos nt So, mx
n =1
( n odd ) ()
! n =1
The total solution is
x t = xh t + " xcn t
From equation (3.33),
()
( ) ( n odd )
1/ 2
2 2 ## 2 & + # 2)! n! & 2 & ! " n ! %% n T n T ' ( ( ' $ $$ ' + 2)! n! . 0.6325n * n = tan "1 - 2 n 2 T2 0 = tan "1 ( ) 10 " n2 , ! n " n !T /
xcn t =
()
an / m
cos n! T t " * n
xcn t =
()
"0.00811
4 2 & n2 # $ n " 19.6 n + 100 ' 1/ 2
# + 0.6325n . & cos % nt " tan "1 ( , 10 " n2 0 /' $
3- 48 So,
x t = Ae
() ()
!" n t
sin " d t # $ + &
n =1
! t = #!" n Ae#!" n t sin " d t # $ x
#!" n t
' ) ' ) #0.00811 #1 + 0.6325n . 2 1 cos 1 nt # tan 2 1/ 2 1 2' 4 2 , 10 # n2 0 / *2 ) ( n n # 19.6 n + 100 1 2 * ( ( *
( n odd )
' ) 0.00811 #1 0.6325n 2 1 + " d Ae cos " d t # $ + & sin nt # tan ( n odd) 1/ 2 2 2 1 ' 4 2 10 # n n =1 ) 1 2 ( n ( n # 19.6 n + 100 * * ' ) % ' ) #0.00811 #1 + 0.6325n . 2 1 x 0 = 0 = # A sin $ + & cos 1 nt # tan n odd 1/ 2 2 0 22 1 2' 4 2 , / 10 # n n =1 ) ( *2 1 ( n ( n # 19.6 n + 100 * *
()
0 = ! A sin " ! 0.00110 ! 0 = 0 = #$ n A sin " + $ d A cos" x ' ) !0.000569 + , +& 1/ 2 + 4 2 2 n =1 ' ) ' ), + ( ( n ! 19.6 n + 100 * (0.00493n + 1* , * 0 = #$ n A sin " + $ d A cos" ! 0.001186
%
()
( n odd )
So A = 0.00117 m and % = - 1.232 rad. The total solution is:
x t = 0.00117 e!0.316t sin 3.15t + 1.23 +#
n =1 "
()
) ( n odd )
$ & $ & !0.00811 !1 ( 0.6325n + / . cos . nt ! tan * m 1/ 2 2 - // . 2$ 4 2 ) , 10 ! n & % '/ . % n % n ! 19.6 n + 100 ' '
3- 49 3.32 Calculate the total response of the system of Example 3.3.2 for the case of a base motion driving frequency of &b = 3.162 rad/s. Solution: Let &b = 3.162 rad/s. From Example 3.3.2,
F t = cY! b cos ! bt + kY sin ! bt = 1.581cos 3.162t + 50sin 3.162t
Also,
()
!n =
k c = 31.62 rad/s and " = = 0.158 m 2 km
! d = ! n 1 # " 2 = 31.22 rad/s
The solution is
2 % 2 # + 2 "# !"# t n b x t = Ae n sin # d t + $ + # nY ' 2 ' 2 2 ' & # n ! # b + 2"# n# b x t = Ae!5t sin 31.22t + $ + 0.0505cos 3.162t ! +1 ! +2
() ()
) (
) )
( * cos # t ! + ! + b 1 2 2 * * )
1/ 2
, 2"# # / +1 = tan !1 . 2 n b = 0.0319 rad 21 - #n ! #b 0 , # / +2 = tan !1 . n 1 = 1.54 rad - 2"# b 0
So, x t = Ae!5t sin 31.22t + " + 0.0505cos 3.162t ! 1.57
!5t !5t
() ( ) ( ) ! ( t ) = !5 Ae sin ( 31.22t + " ) + 31.22 Ae cos ( 31.22t + " ) ! 0.16sin ( 3.162t ! 1.57 ) x # x ( 0 ) = 0.01 = A sin " + 0.0505( 0 ) ! ( 0 ) = 3 ! 5 A sin " + 31.22 A cos" + 0.16 (1) #x
So, A = 0.0932 m and ! = 0.107 rad The total solution is
x t = 0.0932e!5t sin 31.22t + 0.107 + 0.0505cos 3.162t ! 1.57 m
()
