Lecture 15: Coupled Oscillators & Normal Modes
Lecture 15: Coupled Oscillators & Normal Modes • relevant reading: Morin 4.5 • Reminder: Fall break next Monday November 12 & Tuesday November 13 so no tutorials next week. But tutorial questions still available. TA’s will have office hours during their Wednesday tutorial time.
Coupled Systems • So far we have considered systems having only 1 free oscillation (i.e. there is only 1 object that can move independently). This means that the system has only a single natural frequency “ω0 ” associated with it. • ω0 depended on pexample, for p the restoring force in the system. For springs: ω0 = k/m, for small-angle pendulums, ω0 = g/l. • Real physical systems are generally capable of oscillating in many different ways, they may have more than 1 restoring force acting, they may be coupled to other objects and hence affected by their motion, and they may have more than 1 natural frequency. • However, there are still certain circumstances where the components of the system oscillate at a single “natural frequency” and we would like to figure out what this is. These frequencies are known as “characteristic frequencies” or “normal modes” of the system. Once you know the normal modes, it turns out you can write the general solution for the system in terms of the normal modes.
Example: Two Coupled Pendulums • Show demonstration of coupled pendulums.
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• Set-up and approximations: – Take 2 pendulums and a spring. – Make the width between the pendulums the exact width of the unstretched spring (which we will call L), so when the pendulums are vertical, there is no force due to the spring. – Assume the displacements of the pendulum bobs are small and hence that you can make the small angle approximation. – In the small angle approximation, sin θ ≈ tan θ ≈ θ, cos θ ≈ 1. – In the small angle approximation, the vertical motion will be much smaller than the horizontal motion, so assume there is no acceleration in the vertical direction and just consider the motion in the HORIZONTAL “x” direction. (This can be proved by using Taylor expansions for cos θ and sin θ. The displacement in the x direction is: ∆x = l sin θ ≈ lθ. The displacement in the y direction is: ∆y = l(1 − cos θ) ≈ l(1 − (1 − θ2 /2)) = lθ2 /2. Since θ is small, θ2 xA , the spring pulls opposite. So the force from the spring on xB is k(xA − xB ) on xA is −k(xA − xB ). The equations of motion are then mg d 2 xA =− xA − k(xA − xB ), 2 dt l d2 xB mg m 2 =− xB + k(xA − xB ). dt l m
• Notice the coupling term (the term that has both xA and xB ) in each equation. • This is very different from our previous examples above and from previous lectures. The equations of motion for each mass are coupled. Changing xA not only affects A, it also affects B. So we need to figure out how to solve this COUPLED set of differential equations. • Also keep in mind that we are searching for the natural frequencies of the system. This means we are looking for solutions in which the pendulum bobs have the SAME frequency. I am therefore going to use our standard trick of assuming the solution has the form xA = Aeiωt ) 5
and xB = Beiωt ), (notice I gave them the same frequency ω but different amplitudes). I will plug these into the equations, and see if I get any useful information about ω or the constants A and B. Here’s what you get: mg iωt Ae − k(A − B)eiωt l mg iωt Be + k(A − B)eiωt =− l
−mω 2 Aeiωt = − −mω 2 Beiωt
• This is true for all t, so the coefficients of eiωt satisfy a pair of equations for A and B. • We will simplify the notation a bit by using the definitions: ω02 = g/l and ωc2 = k/m (where the c stands for “coupling”). Removing the common factor eiωt we get −ω 2 A = −ω02 A − ωc2 (A − B)
(1)
−ω 2 B = −ω02 B + ωc2 (A − B).
(2)
• This represents two equations in three unknowns: A, B, ω. So we will only be able to find families of solutions (i.e. 1 constant will remain undetermined). • I am going to show 2 methods to solve this set of equations. The methods are IDENTICAL in principle, just different in notation. First: You can solve by substitution • Solving for B in equation (1): B = (ω02 + ωc2 − ω 2 )A/ωc2
(3)
• Solving for A in equation (2): A = (ω02 + ωc2 − ω 2 )B/ωc2
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(4)
• Plugging equation (3) into equation (4) and multiplying through by ωc4 ωc4 A = (ω02 + ωc2 − ω 2 )2 A • This can be rearranged as: 4 ωc − (ω02 + ωc2 − ω 2 )2 A = 0 • Since we don’t want A = 0, the term in brackets must be 0. We therefore need to solve for ω in: 4 ωc − (ω02 + ωc2 − ω 2 )2 = 0 • This can be factored since its a difference of squares (i.e. its of the form a2 − b2 = (a − b)(a + b) = 0) to give: (ωc2 − ω02 − ωc2 + ω 2 )(ωc2 + ωc2 + ω02 − ω 2 ) = 0 • This leaves us with the equation: (ω 2 − ω02 )(ω 2 − 2ωc2 − ω02 ) = 0. • This “characteristic equation” for ω 2 has 2 possible solutions. Either the first term is 0, or the second term is 0. • If the first term is 0 then: ω 2 = ω02 • If the second term is 0 then: ω 2 = ω02 + 2ωc2 • These 2 solutions were identical to the solutions we intuited initially. These are the “natural frequencies” of the system. Next time: we will use these natural frequencies to determine solutions for A and B. We will investigate the other method of solving for the natural frequencies, and look at something called “normal coordinates”.
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Coupled Systems • So far we have considered systems having only 1 free oscillation (i.e. there is only 1 object that can move independently). This means that the system has only a single natural frequency “ω0 ” associated with it. • ω0 depended on pexample, for p the restoring force in the system. For springs: ω0 = k/m, for small-angle pendulums, ω0 = g/l. • Real physical systems are generally capable of oscillating in many different ways, they may have more than 1 restoring force acting, they may be coupled to other objects and hence affected by their motion, and they may have more than 1 natural frequency. • However, there are still certain circumstances where the components of the system oscillate at a single “natural frequency” and we would like to figure out what this is. These frequencies are known as “characteristic frequencies” or “normal modes” of the system. Once you know the normal modes, it turns out you can write the general solution for the system in terms of the normal modes.
Example: Two Coupled Pendulums • Show demonstration of coupled pendulums.
1
• Set-up and approximations: – Take 2 pendulums and a spring. – Make the width between the pendulums the exact width of the unstretched spring (which we will call L), so when the pendulums are vertical, there is no force due to the spring. – Assume the displacements of the pendulum bobs are small and hence that you can make the small angle approximation. – In the small angle approximation, sin θ ≈ tan θ ≈ θ, cos θ ≈ 1. – In the small angle approximation, the vertical motion will be much smaller than the horizontal motion, so assume there is no acceleration in the vertical direction and just consider the motion in the HORIZONTAL “x” direction. (This can be proved by using Taylor expansions for cos θ and sin θ. The displacement in the x direction is: ∆x = l sin θ ≈ lθ. The displacement in the y direction is: ∆y = l(1 − cos θ) ≈ l(1 − (1 − θ2 /2)) = lθ2 /2. Since θ is small, θ2 xA , the spring pulls opposite. So the force from the spring on xB is k(xA − xB ) on xA is −k(xA − xB ). The equations of motion are then mg d 2 xA =− xA − k(xA − xB ), 2 dt l d2 xB mg m 2 =− xB + k(xA − xB ). dt l m
• Notice the coupling term (the term that has both xA and xB ) in each equation. • This is very different from our previous examples above and from previous lectures. The equations of motion for each mass are coupled. Changing xA not only affects A, it also affects B. So we need to figure out how to solve this COUPLED set of differential equations. • Also keep in mind that we are searching for the natural frequencies of the system. This means we are looking for solutions in which the pendulum bobs have the SAME frequency. I am therefore going to use our standard trick of assuming the solution has the form xA = Aeiωt ) 5
and xB = Beiωt ), (notice I gave them the same frequency ω but different amplitudes). I will plug these into the equations, and see if I get any useful information about ω or the constants A and B. Here’s what you get: mg iωt Ae − k(A − B)eiωt l mg iωt Be + k(A − B)eiωt =− l
−mω 2 Aeiωt = − −mω 2 Beiωt
• This is true for all t, so the coefficients of eiωt satisfy a pair of equations for A and B. • We will simplify the notation a bit by using the definitions: ω02 = g/l and ωc2 = k/m (where the c stands for “coupling”). Removing the common factor eiωt we get −ω 2 A = −ω02 A − ωc2 (A − B)
(1)
−ω 2 B = −ω02 B + ωc2 (A − B).
(2)
• This represents two equations in three unknowns: A, B, ω. So we will only be able to find families of solutions (i.e. 1 constant will remain undetermined). • I am going to show 2 methods to solve this set of equations. The methods are IDENTICAL in principle, just different in notation. First: You can solve by substitution • Solving for B in equation (1): B = (ω02 + ωc2 − ω 2 )A/ωc2
(3)
• Solving for A in equation (2): A = (ω02 + ωc2 − ω 2 )B/ωc2
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(4)
• Plugging equation (3) into equation (4) and multiplying through by ωc4 ωc4 A = (ω02 + ωc2 − ω 2 )2 A • This can be rearranged as: 4 ωc − (ω02 + ωc2 − ω 2 )2 A = 0 • Since we don’t want A = 0, the term in brackets must be 0. We therefore need to solve for ω in: 4 ωc − (ω02 + ωc2 − ω 2 )2 = 0 • This can be factored since its a difference of squares (i.e. its of the form a2 − b2 = (a − b)(a + b) = 0) to give: (ωc2 − ω02 − ωc2 + ω 2 )(ωc2 + ωc2 + ω02 − ω 2 ) = 0 • This leaves us with the equation: (ω 2 − ω02 )(ω 2 − 2ωc2 − ω02 ) = 0. • This “characteristic equation” for ω 2 has 2 possible solutions. Either the first term is 0, or the second term is 0. • If the first term is 0 then: ω 2 = ω02 • If the second term is 0 then: ω 2 = ω02 + 2ωc2 • These 2 solutions were identical to the solutions we intuited initially. These are the “natural frequencies” of the system. Next time: we will use these natural frequencies to determine solutions for A and B. We will investigate the other method of solving for the natural frequencies, and look at something called “normal coordinates”.
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