HW 6_solution

n 1 n 1

n 2 ) sin(n x / l ) f ( x, t ) l n 1 n 1 We then multiply the above equation by sin(m x / l ) and integrate from 0 to l , l l l m m n m sin( x ) a ( t )sin( n x / l ) dx sin( x ) an (t )( ) 2 sin( n x / l )dx sin( x) f ( x, t ) dx n 0 l 0 0 l l l n 1 n 1 Due to the orthogonality, we obtain l l l m 2 m m ( a x) f ( x, t )dx ) am sin( 0 l 2 2 l l m m In this problem, we have f ( x, t ) f ( x x0 ) cos t . Then, sin( x) f ( x, t )dx sin x0 cos t . 0 l l Then, we have m 2 2 m m ( a ) am sin( x0 ) cos t l l l We now assume am am cos t , and obtain m 2 2 m ) am sin( x0 ) am 2 ( l l l which yields 2 m x0 ) sin( l l am m 2 ( ) 2 l Finally, the solution is given as 2 n sin( x0 ) l l y ( x, t ) an (t )un ( x) un ( x) cos t . n 2 2 n 1 n 1 ( ) l
(2n 1) 2l 1 3183 rad/s 2 9549 rad/s 1 15,915 rad/s 1 22, 281 rad/s
G

Problem 3. Under the given boundary conditions, the modes can be solved as un ( x) sin(n x / l ) We then assume that the solution can be expressed as the summation of modes, i.e.,
l
am (t )
m (0) a
m

sin m t
4v 1 sin m t . (2m 1) m
The complete solution is given,
( x, t )
4v 1 (2n 1) x sin sin(n t ) 2 n 1 (2 n 1) n
n (t )sin(n x / l ) an (t )( a

or,
y ( x, t ) an (t )un ( x)
n 1

We plug the above expression into the actual equation of motion y y xx f ( x, t ) , and obtain
n (t )un ( x) an (t )un xx f ( x, t ) a
l 1 (2m 1) x m (0) v a cos 2 (2m 1) 2l 2l 0 Hence,
l
ห้องสมุดไป่ตู้
2v 2l (2m 1) x 4v m (0) a cos l (2m 1) 2l (2m 1) 0
m (0) Under the initial conditions of am (0) 0 and a 4v given above, we can obtain (2m 1)
Problem 2. The equation of motion for the torsional vibration is G xx Using the separation of variable, the solution takes the form as ( x, t ) u ( x)T (t ) Substituting it into the governing equation, (t ) ( G )u ( x)T (t ) u ( x)T xx
ME 5420 Mechanical Vibrations Fall 2014 Homework #6 Problem 1. The equation of motion is
( x, t ) EA ( x, t ) M xx (To avoid confusion, here we use M to represent the mass density.) The initial conditions are given as ( x,0) v ( x,0) 0 and Notice that the right end of the bar remains attached to the wall during the impact. The beam is considered as a fixed-free bar (i.e., fixed at the right end, free at the left end; spatial coordinate x is pointing to the left). Therefore we can solve the natural frequencies and vibration modes as EA (2n 1) (2n 1) x n n , and n ; n ( x) sin 2l 2l M We then express the response as (applying modal transform), (2n 1) x ( x, t ) an (t )sin 2l n 1 Plugging the above expression into the original partial differential equation, and multiplying the entire (2m 1) x equation by m ( x) sin , and integrating over the length, based on the mode orthogonality we 2l can derive 1 1 (2m 1) 2 m EA[ m 1, 2,3, Ma ] am 0 , 2 2 2l or 2 m m a am 0 , m 1, 2,3, We now identify the initial conditions for am . Indeed, we can find that ( x,0) 0 am (0) 0 , m 1, 2,3, From ( x,0) v , we have
(0)sin a
n 1 n

(2n 1) x v 2l
(2m 1) x and integrating over the entire length (i.e., again, 2l utilizing the mode orthogonality), we have l l (2m 1) x (2m 1) x m (0)sin 2 dx v sin dx 0 a 0 2l 2l Therefore we may have, Multiplying the above equation by sin
where 2 2

G
. We can then have
u ( x) a1 cos x a2 sin x
Notice that u (0) 0 . We obtain u ( x) a2 sin x At x l we have boundary condition u x ( x) 0 . Therefore, a2 cos l 0 Hence, G (2n 1) , n n 2l So, n

(t ) G u ( x) T ( ) xx 2 u ( x) T (t ) This equation yields two ordinary differential equations, 2T 0 T u xx u 0
which can be rewritten as