21.55 Worksheet 7 - preparation problems - question 1:

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1 Dynamics 76. Worksheet 7 - preparation problems - question : A coupled oscillator with two masses m and positions x (t) and x (t) is described by the following equations of motion: ẍ x + 8x ẍ x +x A. Write down the equations of motion in the form ẍ + Ax, and find its general solution using a substitution of the form x p cos(ωt φ). B. Find the solution if at t particle is displaced distance δ from equilibrium, and was momentarily at rest, while particle was then at equilibrium, at rest. C. Find the normal coordinates in terms of the original ones (remember to normalize your eigenvectors). D. Write down the expression for the kinetic and potential energy in terms of the normal coordinates. Verify that the total energy of each normal mode is separately conserved. Solution: A. The equation is: ẍ ẍ 8 + A x x Let us substitute x p cos(ωt φ). The decond derivative yields: x ω p cos(ωt φ), so the equation becomes: ω 8 p ω cos(ωt φ) (4) p A ω I Denoting ω λ, we can find the eigenvalues λ j of A, by looking for the roots of the determinant of A λi. The condition is: λ 8 λ so ( λ)( λ) + 4 which yields: 4 λ + λ namely λ, ± 6 ± 4 and. The corresponding values of ω are ω, and ω. We need to solve (4), or: Ap (j) p (j) λ j by finding the eigenvectors corresponding to the two eigenvalues λ j. For λ 4 we get: 6 8 p () 9 p () which yields p () p () or p () p () p () / / where we have already normalized the eigenvector. For λ we get: 9 8 p () 6 p () (4) (4) which yields p () p () or p () p () p () / /

2 Dynamics 77 Therefore the general solution is x / β / x cos(t φ )+β / / cos(t φ ) B. Find the solution if at t particle is displaced distance δ from equilibrium, and was momentarily at rest, while particle was then at equilibrium, at rest. The initial condition on the position is: δ / β / This implies: cos( φ )+β / / cos( φ ) (i) β c +β c δ (44) (ii) β c + β c (4) To implement the initial conditions on the velocity we need the derivative of the solution: ẋ / β / / sin( φ ) β / sin( φ ) which imply ẋ (iii) (iv) 6β s +β s (46) β s + β s (47) The latter two imply β s and β s. Since the equations (i) and (ii) would not be satisfied for β i it follows that s i for for i,. So now c i for both i, and we get from equations (i) and (ii): (i) β +β δ (48) (ii) β + β (49) Now from (ii) we get: β β so from (i): β β δ β δ, β δ and the solution becomes: x x δ cos(t) δ cos(t) (4) C. The diagonalizing matrix is: The normal coordinates are: P / / / / x P x such that the eigenvalue problem we solved is P AP ΛwhereΛ to solve / / / / which yields: P A C / / / / B D λ λ. To invert the matrix we need

3 Dynamics 78 so the normal coordinates are x () (x x ), x () ( x +x ) Recall that our solution is: We see that the normal coordinates: x x cos(t) cos(t) δ cos(t) cos(t) x () (x x ) δ cos(t) x () ( x +x ) δ cos(t) do indeed oscillate as a simple harmonic oscillator with the corresponding frequencies. D. The expressions for the kinetic and potential energies in a given normal mode (j) are: So Summing the two we get: V j λ j x (j), Tj (j) x V 4 δ cos(t) δ cos (t) T δ sin(t) δ sin (t) E δ which is indeed constant. For the second normal mode: Summing the two we get: V δ δ cos(t) cos (t) T δ δ sin(t) sin (t) E δ which is conserved as well, as expected.

4 Dynamics 79.6 Worksheet 7 - Workshop problems - question : Longitudinal Oscillations: Consider an elastic string of unstretched length 4a, which is stretched horizontally on a smooth surface between two fixed points at a distance 4a apart (a >a ). There particles of mass m are attached so as to divide the string into four equal sections. We enumerate the segments from left to right, i through 4. The tension T i in each segment i is proportional to its extension (a a ), with the elastic constant being c>. Suppose that the particles perform longitudinal horizontal oscillations (they only move along the line connecting the two fixed points). Assume that these oscillations are sufficiently small so that the strings are never slack: the displacements x i from equilibrium are such that x i < (a a ) for any i,,. A. Show that the tension of the string in segment i is T i c a a + x i x i Write down the explicit expressions for T i with i through 4 (and take proper care of the boundary conditions at both ends!) B. Write the equations of motion for the three particles as differential equations for x i (t). Are these equations homogeneous? C. Bring your equations of motion to a standard matrix form ẍ+n Gx and find n and the matrix G explicitly. D. Look for a normal mode solution using the substitution: x p α cos(ωt)+βsin(ωt) Show that the problem translates into an eigenvalue problem. Find the eigenvalues and the corresponding eigenvectors. Write down the general solution. Solution: A. Show that the tension of the string in segment i is T i c a a + x i x i Let us introduce y i as the coordinates for each mass m i measured from the left. For the leftmost segment i, the tension (acting on body m )is: T c(y a ) for the second (acting on the mass to its right m )itis: For the third (acting on the mass to its right m )itis T c(y y a ) T c(y y a ) and finally for the fourth segment, the tension acting on the right wall is; T 4 c(4a y a )

5 Dynamics 8 The net forces on the masses are: F () total T T c(y a )+c(y y a ) c(y y ) F () total T T c(y y a )+c(y y a ) c(y y y ) F () total T T 4 c(y y a )+c(4a y a ) c(y y 4a) The equilibrium point is such that for all three masses, the net force is zero. Solving the set of three equations F (i) total we get y a, y a and y a, as intuitively expected. We are told that x i are measured with respect to equilibrium, so x y a, x y a and x y a. Thus in terms of x i the tensions are: T c(x + a a ) T c(x x + a a ) T c(x x + a a ) T 4 c(a a x ) which is consistent with T i c a a + x i x i if we assume that the boundary conditions are: x x. B. Write the equations of motion for the three particles as differential equations for x i (t). Are these equations homogeneous? In terms of x i the next forces on the three bodies are: So the corresponding equations of motion are: F () total T T c(y y ) c(x x ) F () total T T c(y y y ) c(x x x ) F () total T T 4 c(y y 4a) c(x x ) mẍ c(x x ) mẍ c(x x x ) mẍ c(x x ) (4a) (4b) (4c) These equations are homogeneous. This is so because x i have been defined with respect to the equilibrium point, so when they all vanish, the forces (and accelerations) are zero. C. Bring your equations of motion to a standard matrix form ẍ+n Gx and find n and the matrix G explicitly. To get the required for we move the force term in (4a) to the lhs, divide by m and define n c/m. We then get: ẍ ẍ + n x x (4) ẍ x D. Look for a normal mode solution using the substitution: x p α cos(ωt)+βsin(ωt) Show that the problem translates into an eigenvalue problem. Find the eigenvalues and the corresponding eigenvectors. Write down the general solution. Let us substitute the suggested solution into (4). We get: ω + n p p p p α cos(ωt)+βsin(ωt) (4)

6 Dynamics 8 Clearly, for this to hold for any t, the term in the square brackets must vanish.s Defining w n λ, this takes the form: p p p p λ (44) which is an eigenvalue problem: we need to determine the eigenvalues λ i and the corresponding eigenvectors p. To this end we need the determinant to vanish: λ λ λ This yields: and therefore ( λ) λ λ + λ ( λ) ( λ) ( λ)( 4λ + λ ) so the eigenvalues are λ and λ + and λ. Let us now compute the eigenvectors: for λ : the eigenvector is: For λ + we get: so p p then Finally for λ we get: so p p then So the general solution is: x x x + + p () p () p () p p p () p () p () p p p p p () p () p () α cos( nt)+β sin( nt) α cos( + nt)+β sin( + nt) α cos( nt)+β sin( nt)

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