HW2, EGME 431 (Mechanical Vibration) Date due and handed in March 9,2008

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HW2, EGME 431 (Mechanical Vibration)
Date due and handed in March 9,2008

Nasser M. Abbasi

June 17, 2014

1 Problem 2.7
1.1 Part(a)
1.2 Part(b)
2 Problem 2.10
3 Problem 2.29
4 Problem 2.46
5 Problem 2.57
5.1 Part(a)
5.2 Part(b)

1 Problem 2.7

Solution sketch: Obtain the Lagrangian, ﬁnd EQM, solve in terms of general initial conditions $x(0) = x0$ and $x_(0) = v0$ , then solve parts (a) and (b) using this general solution.

This is one degree of freedom system. Using $x$ as the generalized coordinates, we ﬁrst obtain the Lagrangian $L$

L = T − U

Where

Hence the EQM is (using the Lagrangian equation), and $F = 10$ and $ω = 10$ rad/sec. (the forcing frequency)

Where $2 -k ω n = m$ . The solution is

x(t)= xh(t) +xp(t)

(2)

To obtain $xht(t)$

¨xh(t) + ω2nxh(t) = 0

Assume $xh(t)= eλt$ and substitute in the above ODE we obtain the characteristic equation

Hence

xh(t)= A cosωnt+ B sinωnt

(3)

Guess

Notice, the above guess is valid only under the condition that $ω ⁄= ωn$ which is the case in this problem. Now, substitute the above 3 equations into (1) we obtain

By comparing coeﬃcients, we obtain

and $c1 = 0$ , hence

|----------------------| | | | xp(t)= --F2∕m2-sinωt | | (ωn−ω) | -----------------------

Then from (2) we obtain

Using (3) in the above

F ∕m x(t) = Acosωnt +B sin ωnt+ (ω2−-ω2-)sin ωt n

(4)

Now assume $x(0 )= x0$ and $_x(0)= v0$ For the condition $x(0) = x0$ we obtain

|--------| | | | x0 = A | ----------

For the condition $_x(0)= v 0$ we obtain

Hence

|--------------------| | | | B = vω0− ωω--F2∕m2- | | n n(ωn−ω ) | ---------------------

Hence (4) can be written as

( ) -v0- ω----F-∕m--- ---F∕m--- x (t)= x0cosωnt + ωn − ωn (ω2n − ω2 ) sinωnt+ (ω2n − ω2)sinωt

Let $ω-= r ωn$ , the above becomes

( ) v0- ---Fm-r---- ---F∕m---- x(t)= x0cosωnt+ ωn − ω2 (1 − r2) sin ωnt+ ω2 (1 − r2) sinωt n n

But $ω2 = k n m$ hence

( ) v0- ---Fm-r--- ----Fm---- x(t)= x0cosωnt+ ωn − k (1 − r2) sin ωnt+ k (1− r2) sinωt m m

Therefore, the general solution is

( ) x(t)= x0cosωnt+ v0− F----r--- sin ωnt+ F----1---sin ωt ωn k (1 − r2) k (1− r2)

(5)

1.1 Part(a)

When $x0 = 0,v0 = 0$ we obtain from (5)

( F r ) F 1 x(t)= − -------2- sinωnt+ -------2-sinωt k (1− r ) k(1− r )

(6)

Substitute numerical values, and plot the solution. $F = 10,ω = 10$ rad/sec. $∘ 2000- ω 10 ,k = 2000,m = 100,ωn = 100-= 4.4721,r = ωn = 4.4721 = 2.2361,$ then equation (6) becomes

In the following plot, we show the homogeneous solution and the particular solution separately, then show the general solution.

1.2 Part(b)

When $x0 = 0.05$ and $v0 = 0$ we obtain from (5)

( F r ) F 1 x(t)= 0.05 cos ωnt+ − k-(1−-r2) sinωnt+ -k(1−-r2)sinωt

Substitute numerical values found in part(a), then the solution becomes

In the following plot, we show the homogeneous solution and the particular solution separately, then show the general solution.

2 Problem 2.10

Following the approach taken in problem 2.7, the EQM is

2 F0 ¨x+ ω nx = m cos ωt

And $x(t)= xh(t)+ xp(t)$ where $xh(t) = Acosωnt+ B sinωnt$ . For $xp(t),$ guess $xp(t) = c1cosωt+ c2sin ωt$ and following the same steps in problem 2.7, we obtain

( 2 2 ) ( 2 2 ) F0 sin ωt − ω c2+ ωnc2 + cosωt − ω c1+ ωnc1 = m cosωt

Notice that the above guess is valid only under the condition that $ω ⁄= ωn$ . Compare coeﬃcients, we ﬁnd $c2 = 0$ and

c = F0---1--- 1 m ω2n − ω2

Hence

xp(t)= F0 ---1---cosωt m ω2n − ω2

Then, the general solution is

F0 ---1--- x(t)= A cosωnt+ Bsinωnt+ m ω2n − ω2 cosωt

(1)

Let, at $t = 0$ , $x(0) = x0$ , and $x_(0 )= v0$ ,then from (1), we ﬁnd

And since

F0 1 _x(t) = − A ωnsin ωnt+ Bωn cos ωnt− ω- -2----2 sinωt m ωn − ω

Then

Therefore, the general solution is (from (1))

|----------------------------------------------------| | ( ) | | x(t)= x0− F0-21-2 cosωnt+ v0sinωnt+ F0-21-2 cosωt | | m ωn−ω ωn m ωn−ω | ------------------------------------------------------

To make the response oscillate at frequency $ω$ only, we can set $v = 0 0$ to eliminate the $sinω t n$ , and set $F0--1--- x0 = m ωn2−ω2$ to eliminate the $cosωnt$ term. Hence, the initial conditions are

3 Problem 2.29

This is one degree of freedom system. Using $x$ along the inclined surface as the generalized coordinates, we ﬁrst obtain the Lagrangian $L$

We ﬁrst note that $kΔ = mg cos𝜃$ and the mass will lose potential as it slides down the surface. We measure everything from the relaxed position (not the static equilibrium.) This is done to show more clearly that the angle do not aﬀect the solution.

L = T − U

Where

Hence

But $kΔ = mg cos𝜃$ , hence the above reduces to

∂L ---= − kx ∂ x

Hence the EQM is (using the Lagrangian equation)

Where $2 -k ω n = m$ . We see that the angle $𝜃$ is not in the EQM. Hence the solution does not involve $𝜃$ and the oscillation magnitude is not aﬀected by the angle. Intuitively, the reason for this is because the angle eﬀect is already counted for to reach the static equilibrium. Once the system is in static equilibrium, the angle no longer matters as far as the solution is concerned.

4 Problem 2.46

From example 2.4.1, we note the following table

Also, from example 2.4.1, the mass of car 1 is $1007kg$ and the mass of car 2 is $1585kg$ . Hence we write

To ﬁnd the deﬂection of the car, we use equation 2.70 in the book, which is

∘ ---------------- ---1+-(2ξr)2---- X = Y (1 − r2)2+ (2ξr)2

Where $X$ is the magnitude of the steady state deﬂection and $Y$ is the magnitude of the base deﬂection, which is given as $0.01$ meters in the example.

Hence, for each speed, we calculate $ωp$ and then we ﬁnd $∘ ----- ωn = m1k+mp$ and then ﬁnd $r = ωωpn$ and then ﬁnd $ξ = -∘--c---- 2 k(m1+mp)$ and then using equation(1), we calculate $X$ . This is done for each diﬀerent speed (all for car $m1 ).$ Next, we do the same for car $m2$ . These calculation are shown in the following table. Note also that $c = 2000$ N s/m as given in the example and $k = 4 ×104$ N/m

car 1


$v (km∕h )$	$ω = 0.2909 v p$	$ω = ∘ --k-- n m1+mp$	$r = ωp ωn$	$ξ = -∘--c---- 2 k(m1+mp)$	$X = Y∘ ---1+(2ξr)2--(cm ) (1−r2)2+(2ξr)2$

$20$	$5.818$	$5.7567$	$1.0106$	$0.14392$	$3.571$

$80$	$23.272$	$5.7567$	$4.0426$	$0.14392$	$0.0997$

$100$	$29.09$	$5.7567$	$5.0532$	$0.14392$	$0.0718$

$150$	$43.635$	$5.7567$	$7.5799$	$0.14392$	$0.0425$

car 2


$v (km ∕h)$	$ωp = 0.2909 v$	$∘ ----- ωn = m2+kmp-$	$ω r = ωpn$	$ξ = ∘---c---- 2 k(m2+mp)$	$∘ -------2--- X = Y --1+2(2ξ2r)-2 (cm) (1−r)+(2ξr)$

$20$	$5.818$	$4.7338$	$1.229$	$0.11835$	$1.7726$

$80$	$23.272$	$4.7338$	$4.9161$	$0.11835$	$0.066141$

$100$	$29.09$	$4.7338$	$6.1452$	$0.11835$	$0.04797$

$150$	$43.635$	$4.7338$	$9.2178$	$0.11835$	$0.02857$

Observations: The heavier car (car 2) has smaller defection ( $X$ values) for all speeds. Adding passengers, causes $ωn$ to change. This results in making the deﬂection smaller when passengers are in the car as compared without them. Heaver cars and heavier passenger results in smaller deﬂection values. For the lighter car however, adding the passenger did not result in smaller deﬂection for all speeds. For speed $v = 20$ , adding the passenger caused a larger defection ( $3.19$ vs. $3.571$ ). As car 1 speed became larger, the deﬂection became smaller for both cars.

So, in conclusion: lighter cars have larger deﬂections at bumps, and the faster the car, the smaller the deﬂection.

5 Problem 2.57

Given $3 m = 120 kg,k = 800× 10$ N/M, $c= 500$ kg/s, and mass $m0$ has angular speed of $3000×2π ωr = 60$ $= 100π$ radians per seconds

5.1 Part(a)

The rotating mass will cause a downward force as the result of the centripetal force $m0e ω2sin(ωrt) r$ . Hence the reaction to this force on the machine will be in the upward direction. Hence

2 Fr = m0eω r sin (ωrt)

Hence the machine equation of motion is

By guessing $xp = Xsin(ωrt− 𝜃)$ then we ﬁnd that (The method of undetermined coeﬃcients is used, derivation is show in text book at page 115)

moe--------r2--------- X = m ∘ -----22-------2- (1− r) + (2ξr)

(2)

This is the maximum magnitude of motion in steady state. In the above, $r = ωr ωn$ . Hence to ﬁnd $X$ we substitute the given values in the above expression. We ﬁrst note that we are told that $2 m0eω r = 374N$ , hence $moe = 3ω72r4$ but we found that $ωr = 100 π$ rad/sec, hence