Hw6_Sol

3
PROBLEM 2 (P. 23 on page 375) If the force input to a single degree of freedom vibration system is F(t) = F ei ωt and the steady–state output is X ei ωt , then the frequency response function is Hx|F (ω) = X ei ωt / F ei ωt , which can be expressed as a magnitude and phase: Hx|F ( ω) = | Hx|F ( ω)| e-iφ X (t) = |F | | Hx|F ( ω)|e-iφ • • The Euler formula says ei θ = cos(θ ) + i s i n ( θ ) . Show that i = ei π / 2 and −1 = ei π . Find H X ω ) and H X ω ) the frequency response functions when the response is the velocity or the & |F ( && | F ( acceleration. & ( t) && ( t) X X − iφ HX & | F (ω ) e && | F (ω ) = i ωt = H X Fe F eiω t Generalize your answer by explaining the effect of phase angle when you take the time derivative of the response. Hint: HX & | F (ω ) = SOLUTION The Euler formula says: We use equation {1} as follows:
{2}
Displacement: Acceleration:
% ei ω t dt = u (t ) = ∫ u & dt = ∫ u & u &&( t ) = d %& i ω t u& ( t ) =i ω ue dt
{3}
We substitute {3} in {2} and cancel out the common non-zero exponential term.
We use equation {4} to evaluate the system response due to evaluate the response of the system as follows:
% (ω ) = F0 H (ω ) = F0 u % ( 200) = 5 u tan (φ ) = 1
2
Therefore, the absolute value of the transfer function H( ω) is a hyperbolic function of the driving frequency ω, where:
∞ H (ω ) = 0
for for
ω→0 ω →∞
1
( −ω m)
2 2
2
+ (ω c )
2
⋅ ) ( − 200 ⋅1) + ( 20010
2
2
= 1.2485 × 10− 4 ≅ 0.00
{7}
ωc 20010 ⋅ =− = − 0.05 ⇒ φ = − 2.86o ≈ 0 − rad 2 −ω m 2002 ⋅1
The system response is almost in phase with the applied forcing function, therefore inertia (mass) is the control factor of the response.
% (ω ) + i ω c u % ( ω ) = F0 −ω 2 m u % (ω ) = u H (ω ) = F0 −ω m + i ω c
2
u % (ω ) 1 = 2 F0 −ω m + i ω c 1 = −ω m + i ω c
2
{4}
H (ω ) =
1
( −ω m)
2
2
+ (ω c )
eiθ = cos(θ ) + i sin(θ )
{1}
π π eiπ / 2 = cos( ) + i sin( ) = 0 + i = i 2 2 iπ e = cos(π ) + i sin( π ) = − 1 +0 = −1
The equation of motion for the simple harmonic oscillator subjected to harmonic force is:
Notice that u (t)=X (t) in this present solution. • For the velocity being of the form:
mu && + cu & + k u = F0 e i ω t % u & =u & ei ω t % & iω t u e iω
We use equation {4} to evaluate the system response due to evaluate the response of the system as follows:
% (ω ) = F0 H (ω ) = F0 u % (0.5 ) = 5 u tan (φ ) = 1
•
(
)
{2}
The free body diagram of the structure is simply as follows. Note also that u (t)=x(t)
cu &
mu &&
F(t)
•
The equation of motion is then formulated as follows:
{1}
We will evaluate the response using complex notation, and the response will be the real part of the result, namely:
u ( t ) = Re( F0 H (ω ) ei ω t ) = Re F0 A (ω ) e i (ω t −φ ) = F0 A(ω ) cos (ω t − φ )
{5}
We plot schematically the absolute value of the transfer function vs. the driving frequency ω.
H (ω ) u / F
ω •
At very low frequency, damping is dominant. Find the response magnitude |X| and the approximate phase angle between the force and the response when ω = 0.5 rad /sec. Let F0 = 5.0 N.
Solutions for Homework #6
PROBLEM 1 (P. 22 on page 375) A simple 1-dof system consists of a mass and a damper only, as shown in the figure below. The excitation is the real part of F(t ) = F0 ei ωt , where F0 is real and positive.
&& + c u & = F (t ) = F0 e i ω t mu
{3}
1
•
% (ω ) e For the response of the structure u (t ) = u
iω t
, we substitute in equation {3} to evaluate the
frequency response function. Therefore we have (we cancel out the common non-zero exponential term):
SOLUTION For the single degree of freedom system, the forcing function is of the following form:
F (t ) = Re ( F0 ⋅ ei ω t ) = Re ( F0 cos (ω t ) + i F0 sin ( ωt ) ) = F0 cos ( ωt )
iω ω 2 ω k 1 − + 2 i ξ ω ωn n % ei ω t u && = u &&
= i ω Hu / F ( ω )
{5}
•
For the acceleration being of the form:
•
At high frequency, the system is mass controlled. Find the response magnitude |X| and the approximate phase angle between the force and the response when ω = 200.00 rad /sec.