<p> FORCED VIBRATION</p><p> m˙x˙  fe sin t  kx  cx˙ k c m˙x˙  cx˙  kx  fe sin t</p><p> m 2 f sin t ˙x˙  2 x˙   x  e n n m</p><p> fe sint</p><p>The general solution of a nonhomogeneous equation consists of two parts:</p><p>(1) the general solution of homogeneous equation</p><p>X h exp(nt)sin(dt  ) representing free vibration</p><p>(2) a particular solution of the nonhomogeneous equation</p><p>Assume xp (t)  X sin(t ) as the particular solution which represents the steady-state oscillation (free vibration part diminishes).</p><p> f e c Solution X    arctan( ) (k  m 2 ) 2  (c) 2 k  m 2</p><p>1 free 3 particular 2</p><p>1</p><p>0 0 4 8 12 -1</p><p>-2</p><p>-3</p><p>Solution of homogenous equation (free vibration)</p><p>Particular solution of nonhomogenous solution</p><p>3 forced</p><p>2</p><p>1</p><p>0 0 4 8 12 -1</p><p>-2</p><p>-3</p><p>Complete solution (forced vibration) x(t)  xh (t)  xp (t)</p><p>2 c k  kX 1 2   f     arctan[ n ] e [1 ( )2 ]2  [2 ( )]2  1 ( )2 n n n kX/fe 10</p><p>8</p><p> 6    4   2</p><p>0 /n nn 0 1 2 3 4  180</p><p>   90   </p><p>/n nn 0 0 1 2 3 4</p><p>3 kX These formulas indicate that the nondimensional amplitude and fe  phase  depend on the frequency ratio . n</p><p>The curves can be divided into three regions:    (1)  1 (smaller than and not close to 1) ; (2)  1; and (3) ℏ 1. n n n</p><p> In the region of close to 1, the nondimensional amplitude n becomes very large.</p><p> kX 1   1 at  fe 2 n</p><p>Damping has a great influence at resonance. If damping is absent, the amplitude becomes infinite at resonance in theory.</p><p> kX  Where dose the maximum occur ( )? What is the maximum? fe n</p><p> Beyond  2 , the nondimensional amplitude is smaller than n one, that is, the amplitude is smaller than the static deflection.</p><p>The other thing to notice is that the response x(t) lags behind the ° ° excitation force by a phase angle   90 for   n and   90 for   n . At   n , there is a steep change of phase angle.</p><p>The natural frequency n (internal) and  (external) should not be confused. 4 Machines with Rotating Parts</p><p>Suppose the center of mass is away from the e centre of rotation. The off-centred mass will create an unbalanced force of  me 2 cost in horizontal direction me 2 sint in vertical direction</p><p>(Complex) Frequency Response Function (information only)</p><p>The excitation can be a sinusoidal or co-sinusoidal function of time. Both cases can be represented by a single function of time.</p><p> m˙x˙  cx˙  kx  fe exp(i t) (i  1 )</p><p>The steady-state solution is assumed as x(t)  X exp(it)</p><p>2 (k  ic  m )X exp(it)  fe exp(i t)</p><p> f f X  e x(t)  e exp(it) k  ic  m 2 k  ic  m 2</p><p>The frequency response function is defined as X 1  2 f e k  ic  m</p><p>5 Support Motion and Vibration Isolation</p><p>Sometimes a vibration is initiated by an oscillatory motion, from ground (earthquake) or an adjacent connection, rather than by an oscillatory force. Theses cases can be considered as support motion.</p><p>Equation of motion x m m˙x˙  k(x  y)  c(x˙  y˙ ) k c let z  x  y y m˙z˙  cz˙  kz  m˙y˙ y(t)  Y exp(it)</p><p>Suppose z(t)  Z exp[i(t )] and x(t)  X exp[i(t )]. Z and , and X and  can be found from the above equation.</p><p>2 2 2 X k  (c) 1[2 ( /n )]  2 2 2  2 2 2 Y (k  m )  (c) [1 ( /n ) ] [2 ( /n )]</p><p>3 2 ( /n ) tan  2 2 1 ( /n ) [2 ( /n )]</p><p>6 18 0 10  X/Y 12 0 8</p><p>6 0 Series1</p><p>0 Series2 6 0 1 2 3 4 5 Series3 Series4 4 Series5 Series6 2</p><p>0  / n 0 1 2 2 3 4</p><p>X From the above graph of , one can see for good isolation, k must be Y  X selected such that  2 (then  1). Such a device is called a Y  n vibration isolator.</p><p>7 Transfer Function</p><p>When the excitation is a harmonic function of time, it is easy to find the response. For a general excitation of f (t) , determining the response in the time-domain can be difficult.</p><p>The equation of motion of a mass-spring-damper system under a general excitation is m˙x˙  cx˙  kx  f (t)</p><p>Taking the Laplace transform and assuming the initial response is zero, yields</p><p>1 X (s)  G(s)F(s) G(s)  ms 2  cs  k and</p><p>X (s) G(s)  transfer function. F(s)</p><p>It can be noticed that G(i) is the frequency response function defined in a previous section.</p><p>Once the excitation f (t) is given in a known system, F(s) can be found and X (s) can also be found from the system’s transfer function. By means of the inverse Laplace transform of X (s) , the system’s response to f (t) , x(t) can be determined.</p><p>Example:</p><p>8 Suppose the excitation to a mass-spring-damper system is shown below</p><p> f(t)</p><p> f 0</p><p> t</p><p> a</p><p>  f exp(sa) F(s)  f (t)exp(st)dt  f exp(st)dt  0 1.  0  0 a s</p><p>1 2. G(s)  ms 2  cs  k</p><p> f exp(sa) 3. X (s)  G(s)F(s)  0 s(ms2  cs  k)</p><p>1 -1 f exp(sa) 4. x(t)  L [X (s)]  L [ 0 ]  H (t  a)u(t  a) s(ms2  cs  k) where H is a unit step function and u is the response of a unit step function.</p><p>9</p>