Classical Mechanics - Homework Assignment 7 Alejandro G´omez Espinosa∗ November 15, 2012
Goldstein, Ch.6, 4 Obtain the normal modes of vibration for the double pendulum shown in Figure 1 assuming equals lengths, but not equal masses. Show that when the lower mass is small compared to the upper one, the two resonant frequencies are almost equal. If the pendula are set in motion by pulling the upper mass slightly away from the vertical and then releasing it, show that subsequent motion is such that at regular intervals one pendulum is at rest while the other has its maximum amplitude. This is the familiar phenomenon of beats.
Figure 1: Sketch of problem 14. Double pendulum. Using Figure 1, we can define the coordinate system in the top of the first pendulum with x pointing to the right and y pointing down. Then, the position of each mass is given by: r1 = l sin θ1 x b + l cos θ1 yb r2 = (l sin θ1 − l sin θ2 )b x + (l cos θ1 + l cos θ2 )b y and the velocities: v1 = r˙ 1 = lθ˙1 cos θ1 x b − lθ˙1 sin θ1 yb v2 = r˙ 2 = l(θ˙1 cos θ1 − θ˙2 cos θ2 )b x − l(θ˙1 sin θ1 + θ˙2 sin θ2 )b y ∗
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and the kinetic energy is: T
m1 v12 m2 v22 + 2 2 2 2 ˙ m1 l θ1 m2 l 2 ˙ 2 ˙ 2 + θ1 + θ2 − 2θ˙1 θ˙2 cos(θ1 + θ2 ) 2 2 1 m2 l2 θ˙22 (m1 + m2 )l2 θ˙12 + − m2 l2 θ˙1 θ˙2 ⇐⇒ cos(θ1 + θ2 ) ≈ 1 2 2
= = =
where the last conditions is due to small oscillations. Thus, the kinetic energy matrix is given by: (m1 + m2 )l2 −m2 l2 T = (1) −m2 l2 m2 l2 In addition, the potential energy: V
= −m1 gl cos θ1 − m2 gl (cos θ1 + cos θ2 ) = −(m1 + m2 )gl cos θ1 − m2 gl cos θ2 ≈ −(m1 + m2 )gl − m2 gl
and the potential energy matrix is given by: (m1 + m2 )gl 0 V = 0 m2 gl
(2)
To calculate the resonants frequencies, let’s use equation (6.12) of Goldstein where the determinant det|V − w2 T | = 0 to obtain non-trivial solutions: 2 −m l2 (m1 + m2 )gl 0 2 2 (m1 + m2 )l 2 −w det |V − w T | = −m2 l2 m2 l 2 0 m2 gl (m1 + m2 )(gl − w2 l2 ) m2 w2 l2 = 2 2 2 m2 w l m2 l(g − w l) = m2 (m1 + m2 )(gl − w2 l2 )2 − m22 w4 l4 = m2 (m1 + m2 )g 2 l2 − 2m2 gl3 (m1 + m2 )w2 + m1 m2 l4 w4 (m1 + m2 )g 2 − 2lg(m1 + m2 )w2 + m1 l2 w4 = 0
w
2
= = =
p
4l2 g 2 (m1 + m2 )2 − 4(m1 + m2 )g 2 m1 l2 2m1 l2 p g(m1 + m2 ) ± g (m1 + m2 )2 − m1 (m1 + m2 ) m l r 1 g(m1 + m2 ) m2 1± m1 l m1 + m2 2lg(m1 + m2 ) ±
If m1 >> m2 , then w2 = gl . To find the normal modes, we have to calculate (V − w2 T ) · a = 0. Using Mathematica, we found that the modes are: r m1 + m2 a = a± x b ± a± yb m2 2
To normalize this expression we use a∗± · T · a± = 1: m1 + m2 2 a± m1 + m2 + m2 = 2(m1 + m2 )a2± m2 and the normalized normal mode are: 1 a± = √ 2
1 1 √ x b ± √ yb m2 m1 + m2
From this expression we can see that the lower normal mode represent the mode with the minus sign. The equation of the normal modes of the system are: θ1 = θ2 =
1 p (A cos (w+ t + ψ+ ) + B cos (w− t + ψ− )) 2(m1 + m2 ) 1 √ (A cos (w+ t + ψ+ ) − B cos (w− t + ψ− )) 2m2
where A, B, ψ+ , ψ− are constants. Using the initial conditions from the system when wemove the top mass slightly away from the vertical, we have: θ1 (0) = θ0 ,
θ2 (0) = 0,
θ˙1 (0) = 0,
θ˙2 (0) = 0
Replacing this conditions in the equations of the normal modes: A cos ψ+ + B cos ψ− p = θ0 2(m1 + m2 ) A cos ψ+ − B cos ψ− √ θ2 (0) = =0 2m2 w+ A sin ψ+ − w− B sin ψ− p θ˙1 (0) = − =0 2(m1 + m2 ) w+ A sin ψ+ − w− B sin ψ− p =0 θ˙2 (0) = − 2(m1 + m2 )
θ1 (0) =
For convenience, we can choose ψ+ = ψ− = 0. Then, from the equations above, we found that: √ θ0 m √1 +m2 . Replacing in the normal modes equations: A = B and A = 2 θ1 = θ2 =
θ0 (cos (w+ t) + cos (w− t)) 2r θ0 m1 + m2 (cos (w+ t) − cos (w− t)) 2 m2
To find the equations of beats, is better to define w = w ± ∆w. Replacing this, we found the expressions:
w+ +w− 2
and ∆w =
w+ −w− , 2
where w± =
θ1 = θ0 cos(wt) cos (∆wt) r m1 + m2 θ2 = −θ0 sin(wt) sin (∆wt) m2 The can see that the oscillation modes are out of phase, that is why when the amplitude of θ1 is maximum, θ2 must be zero and viceversa. 3
Goldstein, Ch.6, 12 Two particles move in one dimension at the junction of three springs, as shown in the figure. The springs all have unstretched lengths equals to a, and the force constants and masses are shown. Find the eigenfrequencies and normal modes of the system.
Figure 2: Sketch of problem 12. Two mass points of equal mass m connected to each other. The positions and velocities of the masses, where the origin is at the left wall, are: r1 = a + x1 ⇒ r˙1 = x˙ 1
r2 = 2a + x2 ⇒ r˙2 = x˙ 2
;
Therefore, the kinetic energy of the system is: mx˙ 21 mx˙ 22 T = + = 2 2
m 0 0 m
(3)
In addition, the kinetic energy of the system with three springs is: kx21 3 kx2 1 V = + (x1 − x2 )2 + 2 = 4kx21 − 3kx1 x2 + 4kx22 = 2 2 2 2
4k −3k −3k 4k
Then, computing the determinant: 4k − w2 m −3k det |V − w T | = −3k 4k − w2 m 2 = 4k − w2 m − 9k 2 2
= 16k 2 − 8kmw2 + w4 m2 − 9k 2 8kw2 7k 2 = w4 − + m m2 7k k 2 2 = w − w − =0 m m k 7k w22 = m m To find the normal modes, we perform (V − w2 T )a = 0. Let’s start with w12 : 4k − k −3k a1 =0 ⇒ 3ka1 − 3ka2 = 0 −3k 4k − k a2 1 ⇒ a1 = a2 ⇒ a= 1 w12 =
For w22 :
4k − 7k −3k a1 =0 −3k 4k − 7k a2
⇒
−3ka1 − 3ka2 = 0
⇒ a1 = −a2
⇒
4
a=
1 −1
(4)
Goldstein, Ch.6, 14 Find the expressions for the eigenfrequencies of the following electrical coupled circuit:
Figure 3: Sketch of problem 14. A electrical coupled circuit. Let’s imagine the current in both loops running clockwise. Then, the energy due to the inductors (T) in the circuit is given by: T
L1 Q˙ 21 L2 Q˙ 22 L3 (Q˙ 2 − Q˙ 1 )2 + + 2 2 2 (L1 + L3 ) ˙ 2 (L2 + L3 ) ˙ 2 Q1 + Q2 − L3 Q˙ 2 Q˙ 1 = 2 2 L1 + L3 −L3 = −L3 L2 + L3 =
In addition the energy in the capacitors (V) is: V
Q21 Q2 (Q2 − Q1 )2 + 2 + 2C1 2C2 2C3 1 1 1 − C3 C1 + C3 = 1 1 − C13 C2 + C3 =
Applying the procedure of small oscillations to find the expressions for the eigenfrequencies: 1 + C13 − w2 (L1 + L3 ) − C13 − w2 L3 2 C 1 =0 det |V − w T | = 1 1 2 − C13 − w2 L3 C2 + C3 − w (L2 + L3 )
1 1 + − w2 (L1 + L3 ) C1 C3
2 1 1 1 + − w2 (L2 + L3 ) − + w 2 L3 = 0 C2 C3 C3
By solving this cumbersome equation, we can found the eigenfrequencies of this circuit.
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Goldstein, Ch.6, 17 A plane triatomic molecule consists of equal masses m at vertices of an equilateral triangle of sides a. Assume the molecule is held together by forces that are harmonic for small oscillations and that the force constants are identical and equal to k. Allow motion only in the plane of the molecule. (a) Without writing the equations of motion, justify your reasoning on the number of normal modes of the system and how many of these modes have zero frequency. In total, it must be six normal modes due to each molecule can move in x and y direction. Since the three masses are conected with ”strings”, their motion is constrain. Then, three of this modes have zero frequency because it represents the ”rigid” motion of all the particles in the x direction, in y direction and, the rotation of the three masses. (b) One of the normal modes corresponds to a symmetrical stretching of all three vertices of the molecule. Find the frequency of this mode. Let’s define the position of the three masses: ! √ a 3 3 r1 = x1 x b + y1 yb r2 = (a + x2 )b x + y2 yb r3 = b+ + x3 x a + y yb 2 2 Then, the potential energy is given by: m 0 0 0 0 0 0 m 0 0 0 0 0 0 m 0 0 0 m 2 2 2 2 2 2 T = x˙ + y˙ 1 + x˙ 2 + y˙ 2 + x˙ 3 + y˙ 3 = 2 1 0 0 0 m 0 0 0 0 0 0 m 0 0 0 0 0 0 m In the case of the potential energy, we found: V =
k k k (d12 − a)2 + (d23 − a)2 + (d13 − a)2 2 2 2
where: d212 = (a + x2 − x1 )2 + (y2 − y1 )2 !2 √ a 2 3 d223 = − + x3 − x2 + a + y3 − y2 2 2 !2 √ 2 a 3 d212 = + x3 − x1 + a + y3 − y1 2 2 Expanding this expressions and considering only the lower order terms: d12 = a + (x2 − x1 ) + ... d23 d13
√ 1 3 = a − (x3 − x2 ) + (y3 − y2 ) + .. 2 2 √ 1 3 = a + (x3 − x1 ) + (y3 − y1 ) 2 2
Therefore the potential energy is given by: √ √ 2 k √ √ 2 k k V ≈ (x2 − x1 )2 + x2 − x3 − 3y2 + 3y3 + x3 − x1 − 3y3 + 3y1 2 8 8 6
√ √ 3k/4 −k 0 −k/4 − 3k/4 5k/4 √ √3k/4 3k/4 0 0 − 3k/4 √ −3k/4 √ 0 5k/4 − 3k/4 −k/4 3k/4 −k √ √ = 0 0 − 3k/4 3k/4 3k/4 −3k/4 √ √ −k/4 − 3k/4 √ −k/4 3k/4 k/2 0 √ − 3k/4 −3k/4 3k/4 −3k/4 0 3k/2
Using the matrix above, we can calculate the modes with det |V − w2 T | = 0. With help of Mathematica, the eigenvalue of the symmetrical stretching mode is w2 = 3k m
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