Session-XI (17.5.05) BKS
6.Vibrations of Two Degree of Freedom Systems
6.1 Introduction The modeling method discussed in previous chapters employed only one coordinate to describe the motion of the system completely. But general mechanical systems require several degrees of freedom for a meaningful model. Systems modeled with two independent co-ordinates to describe their motion are called two Degree of Freedom systems. There are two equations of motion for a two DOF system, one for each mass. They are generally in the form of coupled differential equations- i.e., each equation involves all the coordinates. If a harmonic solution is assumed for each co-ordinate, the equations of motion lead to a frequency equation that gives two natural frequencies for the system. If a suitable initial excitation is given the system vibrates at one of these natural frequencies. During free vibrations at one of the natural frequencies, the amplitude of two degrees of freedom (coordinates) are related in a specific manner and the configuration is called principal mode, or normal mode or natural mode of vibration. Thus a two DOF system has two normal modes of vibration corresponding two natural frequencies. 6.2 Free vibrations of two DOF system: Consider a two DOF system as shown in Figure 6.1, executing free vibrations. Let an initial displacement X1 be given to mass m1 and X2 to mass m2. Figure 6.2 shows the corresponding free body diagram.
K1 m1 K2
X1
m2 K3 Figure 6.1
X2
Let X2 > X1 K1 X1
K1 X1 m1
m1 X1
K2 X2
X1
K2 X1
K2 (X2 – X1)
K2 X2
K2 X1
K2 (X2 – X1)
m2
m2 K3 X2
X2 K3 X2
Figure 6.2 Based on Newton’s second law of motion ∑ƒ = mX For mass m1 ..
..
m1x1 = - K1x1 + K2 (x2-x1)
..
m1x1 + K1x1 – K2 x2 + K2x1 = 0
..
m1x1 + x1 (K1 + K2) = K2x2
----- (1)
for mass (2)
..
m2x2 = - K3x2 – K2 (x2 – x1)
.. mx +K .. 2 2
3
x2 + K2 x2 – K2 x1
m2 x2 + x2 (K2 + K3) = K2x1
----- (2)
X2
Let us assume that under steady state conditions the solutions for x1 and x2 be harmonic therefore, assume x1 = X1 sin ωt, x2 = X2 sin ωt
.. x = - ω2X sin ωt, 1 1
.. x = - ω2 X sin ωt 2 2
Substitute these in (1) and (2) - m1ω2X1 sin ωt + (K1 + K2) X1 sin ωt = K2 X2 sin ωt - m2 ω2X2 sin ωt + (K2 + K3) X2 sin ωt = K2 X1 sin ωt. Removing sin ωt through out and re arranging the terms. X1/X2 = K2/(K1 + K2 – m1ω2) = [(K2 + K3) – m2ω2]/K2 Cross multiplying K22 = (K1 + K2 – m1ω2) (K2 + K3 – m2ω2) On simplification we get m1m2 ω4 – [m1 (K2 + K3) + m2 (K1 + K2)] ω2 + [K1K2 + K1K3 + K2K3] = 0 The above equation is quadratic in ω2 and gives two values of ω2 and therefore the two positive values of ω correspond to the two natural frequencies ωn1 and ωn2 of the system. The above equation is called frequency equation since the roots of the above equation give the natural frequencies of the system. Discussions: Let K1 = K3 = K m1 = m2 = m Then the frequency equation becomes m2ω4 – 2 m (K + K2) ω2 + (K2 + 2KK2) = 0 Let: ω2 = Ω ∴Ω2 = ω4, m2 Ω2 – 2 m (K + K2) Ω + (K2 + 2 KK2) = 0 ∴m2 Ω2 – 2 m (K + K2) Ω + (K2 + 2KK2) = 0 The roots of the above equation are as follows: Let a = m2, b = -2 m (K + K2); c = (K2 + 2KK2) ∴Ω1,2 = [- b ± √(b2 – 4ac)]/2a = [- (-2m) (K + K2) ± √[-2m (K+K2)]2 – 4 (m2) (K2 + 2KK2)]/2m2
= [+ 2m (K +K2)]/2m2 ± [√4m2[(K2 + k22 + 2 KK2) – (K2 + 2KK2)]/4m4 = (K+ K2) /m ± √(K22/m2) = (K +K2) /m ± K2/m ∴Ω2 = (K + 2K2) /m ωn22 = (K + 2K2) /m
∴ω n2 = √[(K + 2K2) /m]
Ω1 = (K + K2) /m – K2 / m = K/m ωn12 = K/m ∴ω n1 = √(K/m) ωn1 is called the first or fundamental frequency or I mode frequency, ωn2 is called the second or II mode frequency. Thus the number of natural frequencies of a system is equal to the number of degrees of freedom of system.
Session-XII (18.5.05) BKS Two DOF System (contd.) Modes Shapes: From X1/X2 = K2/(K+K2) -mω2 = (K2 + K) - mω2/K2 Substitute ωn1 in any one of the equation. (X1/X2)ωn1 = K2 / K+ K2 – m . K/m (X1/X2)ωn1 = 1 (X1/X2)ωn2 = K2 / K + K2 – m(K+ 2K2/m) = K2/-K2 = -1 (X1/X2)ωn2 = -1 The displacements X1 and X2 corresponding to the two natural frequency of the system can be plotted as shown in Figure 6.3, which describe the mode in which the masses vibrate. Such a diagram is called principal mode shape of the system. When the system vibrates in principal mode the masses oscillate in such a manner that they reach maximum displacements simultaneously and pass through their equilibrium points simultaneously or all moving parts of the system oscillate in phase with one frequency. Since the ratio X1/X2 is important rather than the amplitudes themselves, it is customary to assign a unit value of amplitude to either X1 or X2. When this is done, the principal mode is referred as normal mode of the system.
m2
m1
K1 X1
X2
m1 K2
I Mode m1
m2
.
X1 K3 Figure 6.3
Node
X2 m2
6.3 Discussion on Natural frequencies and mode shapes: Observation 1: It can be seen from the figure when the system vibrates in first mode, the amplitude of two masses remain same. The motion of both the masses are in phase i.e., the masses move up or down together. When the system vibrates in II mode the displacement of two masses have the same magnitude with opposite signs. Thus the motions of m 1 and m2 are 1800 out of phase. Observation 2: When the system vibrates in first mode, the length of the middle spring remains constant, this spring (coupling spring) is neither stretched nor compressed. It moves bodily with both the masses and hence totally ineffective as shown in Figure 6.4. Even if the coupling spring is removed the two masses will vibrate as 2 SDOF system with ωn = √(K/m). Where as when it vibrates in II mode, the midpoint of the middle spring remains stationary for all the time. Such a point which experiences no vibratory motion is called a node, as shown in Figure 6.5. Observation 3: When the two masses are given equal initial displacements in the same direction and released, they will vibrate in I mode. When they are given equal initial displacements in opposite direction and released they will vibrate in II mode as shown in Figures 6.4 and 6.5
K1
K1 X1
m1
K1
m1 K2
X1
m1
K2 K2 X2
m2
m2 K3
K3
Figure 6.4
X2
m2 K3
K1
K1 X1
m1 K2
N
m2
X2 K3
K1
m1
.
m1
K2
N
.
m2
K2
m2 K3
K3
Figure 6.5 If unequal displacements are given in any direction, the motion will be superposition of two harmonic motions corresponding to the two natural frequencies.
Numerical Example 1. Obtain the frequency equation for the system shown in Figure. Also determine the natural frequencies and mode shapes when K1 = 2K, K2 = K, m1 = m, m2 = 2m. K1 X1
K1 X1
K1 m1 m1
X1 K2
m2
m1
X1 K2 (X2 – X1)
K2 X2
K2 X1
K2 X2
X2
X1
K2 (X2 – X1)
K2 X1 m2
X2
m2
X1
From NSL .. for mass (1) m1X1 = - K1X1 + K2 (X2 – X1) .. = - K1X1 + K2 X2 – K2X1 m1X1 + X1 (K1 + K2) = K2 X2 ----- (1) For .. mass (2) m2X2 = - K2 (X2 – X1) .. = - K2 X2 + K2 X1 m2 X2 + K2X2 = K2 X1
----- (2)
= A Sin ωt X1 = - ω2 A Sin ωt,
X2 = B Sin ωt X2 = - ω2 B Sin ωt
..Let X
1
Substitute these in (1) and (2) -m1ω2 A Sin ωt + (K1 + K2) A Sin ωt = K2 B Sin ωt A (K1 + K2 – m1ω2) = K2B A/B = K2 / [(K1 + K2 – m1ω2)] ----- (3) 2 - m2 ω B Sinωt + K2B Sin ωt = K2 A Sin ωt (K2 – m2ω2) B = K2 A A/B = [K2 – m2ω2] / K2 ----- (4) Equating (3) and (4) K2 / (K1 + K2 – m1ω2) = [K2 – m2ω2] /K2 K22 = (K1 + K2 – m1ω2) (K2 – m2ω2) K22 = (K1 + K2) K2 – m1ω2 K2 – m2ω2 (K1 + K2) + m1 m2ω4 m1 m2 ω4 - ω2 [m1 K2 + m2 (K1 + K2)] + K1 K2 = 0 Put ω2 = Ω m1 m2 Ω2 – Ω [m1 K2 + m2 (K1 + K2)] + K1K2 = 0 Or Ω = [[m1 K2 + m2 (K1 + K2)] ± √ [{m1K2 + m2(K1+K2)}2]- 4 m1 m2K1K2]] / 2m1m2 Frequency equation of the system To determine the natural frequencies Given K1 = 2 K, K2 = K m1 = m, m2 = 2m
Ω = [mK + 2m (2K +K) ± √[mK + 6mK)2 – 4m 2mK2K]] / 2m . 2m = [7 mK ± √[(7mK)2 – 4 (4m2K2)]] / 4m2 = [7mK ± √(49m2K2 – 16m2K2] / 4m2 Ω = [7mK ± 5.744 mK] /4m2 Ω1 = ωn12 = [7 mK – 5.744 mK] /4m2 = 1.255 mK /4m2 = 0.3138 K/m ω n1 = 0.56 √(K/m) Ω2 = ωn22 = [7mK + 5.744 mK] /4m2 = 3.186 K /m ω n2 = 1.784 √(K/m) To determine the mode shapes: I mode shape: Substituting ωn12 = 0.3138 K/m A/B = [K2 – m2 ω2] /K2 = [K2 –m2. ωn12]/K2 A/B = (K – 2m.0.3138 K/m)/K = 1-2(0.3138)] A/B = 0.3724 If A = 1, B = 2.6852 II mode: Substituting ωn22 = 3.186 K/m A/B = [K2 –m2ωn22] /K2 = (K – 2m. 3.186 K/m) / K = (1 – 3.186 *2) = - 5.372 A/B = - 5.372, if A = 1, B = - 0.186
A =1
.
B = 2.6852
A =1
. . . B = -0.186
I Mode
2. Determine the natural frequency and the corresponding mode shapes for the system shown in figure X1
X2
K1
K2
K3
m1 •
m2 •
•
•
Given K1 = 3K, K2 = 2K, K3 = K m1= m, m2 = 2m Free body diagram
X1 K1X1
m1
X2 K2X1
K2X1 K2X2
m2
K3X2
K2X2
K2 (X2 –X1) K1X1 ω n1 = √(K/m)
K2 (X2 –X1) ω n2 = √(5.5 K/m)
K3X2
Session-XIII (20.5.05) BKS Two DOF systems (contd.) 3. Determine the Natural frequencies and ratio of amplitudes of the system shown in Figure. Solution similar to example No. 1 ω n1 = 0.517 √(K/m) (A/B)ωn1 = 0.731
ω n2 = 1.931 √(5.5 K/m) K
(A/B)ωn2 = -0.2732 m
2K
2m
4. Same as above Given m1 = 1.5 kg
m2 = 0.8 kg K1
K1 = K2 = 40 N/m ω n1 = 9.39rad/sec (A/B)ωn1 = -0.765
ω n2 = 3.88 rad/sec
m1
(A/B)ωn2 = 0.696 K2
m2
5. Determine the natural frequencies of the system shown in figure. Also determine the ratio of amplitudes and locate the nodes for each mode of vibration. Assume that the tension ‘T’ in the string remains unchanged, when the masses are displaced normal to the string.
l
l
l
m1
m2
m1 T T x1
T α
x1 - x2
β
m2 T x2 γ
Masses in displaced position
T cos α
x1
T
α T sin α
x1
m1
T cos β
β
T
T (x1 - x2) β T sin β T cos β
T sin β m2
x2 T cos γ
γ T sin γ
Free body diagram NSL. For mass (1)
T
..
mx1 = - T Sin α - T Sinβ
Sin α = x1/l, Sinβ = x2/l (x1 – x2/l
= - T x1 /l – T (x1-x2)/l = - Tx1/l – Tx1/l + Tx2/l
..
∴mx1 + 2Tx1/l = Tx2/l
----- (a)
NSL. For mass (2)
..
mx2 = - T Sinγ + T Sinβ = - Tx2/l + T. (x1 –x2)/l
..
∴mx2 + 2Tx2/l = Tx1/l Let
ωt, ..x1 = A Sin 2
x1 = - ω A Sin ωt
----- (b) ωt .. x2 = B Sin 2
x2 = -ω B Sin ωt,
Substitute in (a) and (b) - m1ω2 A Sin ωt + (2T/l) A Sin ωt = (T/l) B Sin ωt. Removing sin ωt throughout A [(2T/l) – m1ω2)] = B. (T/l) ∴A/B = (T/l)/ [(2T/l) – m1ω2)]
----- (a1)
Similarly - m2 ω2. B Sin ωt + (2TB Sin ωt)/l = (T. A Sin ωt)/l B [(2T/l) – m2ω2)] = A. (T/l) ∴A/B = [(2T/l) – m2ω2)]/T/l
----- (a2)
Equating (a1) and (a2) and cross multiplying (T/l)/ [(2T-lm1ω2)/l] = [(2T – lm2ω2)/l]/(T/l) ∴T2 = (2T – lm1 ω2) (2T – lm2ω2) T2 = 4T2 – 2Tlm1ω2 – 2Tlm2ω2 + l2m1m2ω4 ∴l2m1m2 ω 4 – 2Tl (m1 + m2) ω 2 + 3T2 = 0 Let Ω = ω2
Frequency Equation
∴l2m1m2 Ω2 – 2Tl (m1 + m2) Ω + 3T2 = 0 Ω1, 2 = [2Tl (m1 + m2) ± √[{2T (m1 + m2)l}2 – 4 l2m1 m2 3T2)] / 2 m1 m2 l2 Let = m1 = m2 = m ∴Ω1.2 = [2Tl (m + m) ± √[2T (2ml)2 – 4l2m2. 3T2)] / 2. m2. l2 = 4mTl ± √[(4mTl)2 – 12 m2 l2 T2] / 2m2l2 On further simplification Ω 1 = ω n12 = T/ml ∴ω n1 = √(T/ml) 2 Ω 2 = ω n = 3T/ml ∴ω n2 = √(3T/ml) Mode Shape: A/B = (T/l)/[(2T/l) – m1 ω2] I mode: A/B = 1
if A = 1, B = +1
II mode: A/B = -1 A/B = -1
if A = 1, B = -1
(A/B)ωn1 = 1
m1
m2
X1
X2
I Mode m1
.
X1 Node
Semi Definite Systems or Degenerate System
II Mode
(A/B)ωn1 = -1
m2
Eg: Coupled locomotive
Systems for which one of the natural frequencies is equal to zero are called semi definite systems. X1
X2 K
m1 •
m2 •
•
•
FBD:
x2 > x1 X1 m1
X2 KX2
KX2 KX1
m1
KX1
K (X2 –X1) K (X2 –X1)
Mass..(1) ∴ m1 x1 = K (x2 –x1)
..
m1 x1 + Kx1 = Kx2
----- (1)
..
m2 x2 = - K (x2 – x1)
..
m2 x2 + Kx2 = Kx1 Let
ωt, ..x1 = A Sin 2
x1 = - ω A Sin ωt
----- (2) ωt .. x2 = B Sin 2
m2
x2 = - ω B Sin ωt,
m2
Substitute in (1) and (2) m1 (- ω2 A Sin ωt) + K A Sin ωt = K B Sin ωt Further simplifications leads to A/B = (K)/ [K – m1 ω2]
----- (3)
m2 (- ω2 B Sin ωt) + K B Sin ωt = K A Sin ωt Further simplifications leads to A/B = [K – m2 ω2] / (K)
----- (4)
Cross multiplying and simplifying further m1 m2 ω 4 – K (m1 + m2) ω 2 = 0
Frequency equation
ω2 [m1 m2 ω2 – K (m1 + m2)] = 0 Finding the roots we get the natural frequencies ω 1 = ω n1 = 0 ω 2 = ω n2 = √[{K(m1 + m2)}/(m1 * m2)] When one of the roots of the frequency equation is zero, one of the natural frequencies is zero. Such systems are referred as semi definite systems. The system will move as a rigid body without any distortion of spring. The amplitudes of two masses are equal. They are also referred as free-free system. Mode Shapes: I mode: (A/B)ωn1 = (K)/ [K – m1 ω2] ω n1 = 0
m1 0
m2 0 A =1
B =1
I mode
(A/B)ωn1 = 1
II mode:
0 m1 A=1 Node
. B=-1
II mode
0 m2
(A/B)ωn2 = (K)/ [K – m1 ω2] ω n2 = √[{K(m1 + m2)}/(m1 * m2)] if m1 = m2 =m Then (A/B)ωn2 = -1 6. Determine the natural frequency and mode shapes of the system shown in Figure. Given m1 = 10 kgs, m2 = 15 kgs, K = 320 N/m X1
X2 K
m1 •
m2 •
•
•
Solution: It is a free –free system Free body diagram
x2 > x1 X1 m1
X2 KX2
KX2 KX1
Frequencies
m1
ωn2 = √[K (m1 + m2)/(m1 * m2)] ωn2
= √[{320(10 + 15)}/ (10*15)] = 7.30 rad/sec
Mode Shapes I mode
KX1
K (X2 –X1)
∴ω n1 = 0
m2
K (X2 –X1)
m2
(A1/A2)ωn1 = 1.0, if A = 1, B = 1 m1 0
m2 0 A =1
B =1
I mode II mode (A1/A2)ωn2 = (K)/[K – m1ωn22] = 320 / [320 – 10 * (7.30)2] = - 1.49 if A = 1, B = -0.671 0 m1
.
A=1
Node
0 m2
B = - 0.671
II mode 7. An electric train made of two cars each of mass 2000 kgs is connected by couplings of stiffness equal to 40 * 106 N/m. Determine the natural frequency of the system.
m1
m2 K
Coupled Cars Solution: This is an example similar to problem No. 6 only the answer are given here.
Given m1 =m2 = 2000 kgs. K = 40* 106 N/m ωn1 = 0 ωn2
= √(2K/m) = √(2*40*106) /2000
ωn2 = 200 rad/sec
Analysis of Two DOF Torsional Systems
Figure above shows a two rotor system which can be represented as follows.
J1
Kt
Free body diagram is as given below.
J2
θ2
θ1 Kt θ 2
J1
Kt θ1
J2 Kt θ1
Kt θ 2
θ1
θ2 Kt (θ 1 - θ 2)
Kt (θ 1 - θ 2)
θ2 Kt (θ 1 - θ 2)
θ1
Kt (θ 1 - θ 2)
θ1 and θ2 in CCW direction looking from left. NSL for Rotor (1)
..
J.. 1 θ1 = - Kt (θ1 - θ2) J1 θ 1 + Kt θ 1 = Kt θ 2
----- (1)
For .. rotor (2) J.. 2 θ2 = + Kt (θ1 - θ2) J2 θ2 = + Kt θ1 – Kt θ2
..
J2 θ 2 + Kt θ 2 = Kt θ 1 Let,
----- (2)
ωt, ..θ1 = A sin 2
..θ2 = B sin2 ωt θ1 = - ω A sin ωt, θ2 = - Bω sin ωt Substituting the above in 1 and 2 and simplifying we get the amplitude ratios and frequency equation as follows. A/B = Kt/[Kt – J1ω2] ----- (a1) A/B = [Kt – J2ω2] / Kt
----- (a2)
Frequency equation J1 J2 ω4 – (J1 + J2) Kt ω2 = 0 ω2[J1J2ω2 – (J1 + J2) Kt] = 0 ∴ω 2 = 0,
ω n1 = 0
and or J1 J2 ω2 – (J1 + J2) Kt = 0 ωn22 = [(J1 + J2) Kt] /J1 * J2 ω n2 = √[{(J1 + J2) Kt}/ J1 *J2] 8. Determine the natural frequency of Torsional vibrations of a shaft with two circular disks of uniform thickness at its ends. The masses of the discs are m1 = 500 kgs and m2 = 1000 kgs and their outer diameter D1 = 125 cm and D2 = 190 cm. The length of the shaft is 3 m and its diameter = 10 cm. Modulus of rigidity for shaft material of the shaft G = 0.83 * 1011 N/m2 Also determine in what proportion the natural frequency of the shaft gets changed if along half the length of the shaft the diameter is increased from 10 cm to 20 cm. Solution: Part (1) For free body diagram and expression for frequencies refer previous discussion. m1 = 500 kg m2 = 1000 kg D1 = 1.25 m D2 = 1.9 m l = 3.00 m d = 0.10 m G = 0.83 * 1011 N/m2 Two rotor system is a semi definite system whose natural frequency is given by ωn1 = 0 ω n2 = √[{(J1 + J2) Kt}/ J1 *J2]
ωn = √[Kt(J1 +J2)/J1J2] J1 = ½ m1 R12 = 98 kg – m2,
J2 = ½ m2 R22 = 453 kgm2
Kt = GIp/l = [0.83 * 1011 / 3.00] *[π (d4)/ 32] = 2.725 * 105 N-m/ rad ω n2 = 58.1 rad/sec Part (2): Since the diameters are different along the length equivalent stiffness is to be determined as follows.
J1
Kt1
J2
Kt2
Kte J1
J2
Equivalent System Given, d1 = 10 cm, d2 = 20 cm, l1 = l2 = 1.5 m ∴1/Kte = 1/Kt1 + 1/Kt2 Kte = 5.13 * 105 N-m/rad ω n2 = 79.597 rad/sec ∴Hence there is 37% increase in the natural frequency of the system. 9. Determine the frequency equation, natural frequency and mode shapes for a double pendulum shown in figure.
Given m1 = m2 l1 = l2 = l
l1 m1 l2 m2
Free body diagram
θ1
l1 T1
m1 T2 θ2
x1
l2 T2
m2
m 1g x2 m 2g
T1 T1Cos θ1 θ1 T2Sin θ2 T1Sin θ1
θ2
T2Cos θ2
T2 T2 θ2
m1g T2Cos θ2
T2Sin θ2
m2g
Considering only the oscillation Applying NSL for mass (1)
..
m1 x1 = - T1 sin θ1 + T2 sin θ2
..
..
but x1 =lθ1 ∴x1 = lθ1,..x2 =..lθ1 +..lθ2 x2 = lθ1 + lθ2
..
m1 lθ1 = - T1 sin θ1 + T2 sin θ2 At mass (1)
----- (a)
T1 cosθ1 = mg + T2 cosθ2
----- (1)
At mass (2) T2 cosθ2 = mg θ2 being very small cosθ = 1 T2 = mg
----- (2)
∴T1 = mg + mg T1 = 2mg
..
----- (3)
∴mlθ1 = - 2 mg sin θ1 + mg sin θ2
..
lθ 1 + 2gθ 1 - gθ 2 =0
----- (b)
Similarly for mass (2)
..
= - T2 sin θ2
mx2
T2 cos θ2 = m2 g,
T2 = mg
= - mg sin θ2 = - mgθ2
..
..
ml (θ1 + θ2) + mg θ2 = 0
..
..
lθ 1 + lθ 2 + gθ 2 = 0
----- (c)
Equations (b) and (c) represent GDE Let θ1 = A sin ωt
..
θ1 = - ω A sin ωt, 2
θ2 = B sin ωt
..
θ2 = - B ω2 sin ωt
Substitute in (b) and (c) - lω2 A sin ωt + 2 g A sin ωt – g B sin ωt = 0 A (2g - lω2) = Bg A/B = g/[2g-lω 2]
----- (b1)
-lω2 A sin ωt - lω2 B sin ωt = - g B sin ωl A/B = (lω2 – g)/lω2 A/B = [g - lω 2] / lω 2
----- (c1)
Equating b1 and c1 and cross multiplying we get frequency equation. l2ω 4 – 4glω 2 + 2g2 = 0 Let Ω = ω2 ∴l2 Ω2 – 4glΩ + 2g2 = 0
frequency equation
The roots are Ω1 = 0.5857 g/l = ωn12 ω n1 = 0.7655 √(g/l) Ω2 = ωn22 = 3.414 g/l ∴Ω 2 = ω n2 = 1.847 √(g/l) Mode shapes I mode (A/B)ωn1 = g /[2g - lωn12] = 1/1.4143
A/B = 1/1.4143 ∴A = 1, B = + 1.4143 II mode (A/B)ωn2 = g /[2g - lωn22] = 1/-1.4143
A = 1,
B = -1.414
