wo identitical pendula each of length l and with bobs of mass m are free to oscillate in the same plane. The bobs are joined by a spring with spring constant k, by looking for solutions where x and y vary harmonically at the same angular frequency ω, form a simultaneous equation for the amplitudes of oscillation x 0 and y 0 .Considering the forces acting on each pendulum we can derive the following coupled-differential equations: (1) m x ¨ = k ( y − x ) − m g x ℓ (2) m y ¨ = − k ( y − x ) − m g y ℓ Where x and y are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency ω then we can write ∃ ω , ϕ 1 , ϕ 2 ∈ R: x ( t ) = x 0 cos ⁡ ( ω t + ϕ 1 ) y ( t ) = y 0 cos ⁡ ( ω t + ϕ 2 ) Substituting these solutions back into ( 1 ) and ( 2 ) we get: − m ω 2 x 0 cos ⁡ ( ω t + ϕ 1 ) = k ( x 0 cos ⁡ ( ω t + ϕ 1 ) − y 0 cos ⁡ ( ω t + ϕ 2 ) ) − m g ℓ cos ⁡ ( ω t + ϕ 1 ) − m ω 2 y 0 cos ⁡ ( ω t + ϕ 2 ) = − k ( x 0 cos ⁡ ( ω t + ϕ 1 ) − y 0 cos ⁡ ( ω t + ϕ 2 ) ) − m g ℓ cos ⁡ ( ω t + ϕ 2 ) However, without assuming that ϕ 1 = ϕ 2 , in which case everything factors out nicely to leave a simultaneous equation in x 0 and y 0 , I cannot see a way of making it linear in x 0 and y 0 . So am I expected to use this assumption or is there a mathematical way of simplifying it?If it is the former, then what would the physical justification for this assumption be?

Question

wo identitical pendula each of length   l and with bobs of mass   m are free to oscillate in the same plane. The bobs are joined by a spring with spring constant   k, by looking for solutions where   x and   y vary harmonically at the same angular frequency   ω, form a simultaneous equation for the amplitudes of oscillation       x      0   and       y      0  .Considering the forces acting on each pendulum we can derive the following coupled-differential equations:                    (1)                    m                        x          ¨                                    =        k        (        y        −        x        )        −                              m            g            x                    ℓ                                    (2)                    m                        y          ¨                                    =        −        k        (        y        −        x        )        −                              m            g            y                    ℓ                    Where   x and   y are the displacements of each of the pendulum as functions of time. If we assume they oscillate harmonically with angular frequency   ω then we can write   ∃  ω  ,      ϕ      1    ,      ϕ      2    ∈  R:                    x        (        t        )                            =                  x                  0                cos        ⁡        (        ω        t        +                  ϕ                  1                )                            y        (        t        )                            =                  y                  0                cos        ⁡        (        ω        t        +                  ϕ                  2                )            Substituting these solutions back into   (  1  ) and   (  2  ) we get:                    −        m                  ω                  2                          x                  0                cos        ⁡        (        ω        t        +                  ϕ                  1                )                            =        k        (                  x                  0                cos        ⁡        (        ω        t        +                  ϕ                  1                )        −                  y                  0                cos        ⁡        (        ω        t        +                  ϕ                  2                )        )        −                              m            g                    ℓ                cos        ⁡        (        ω        t        +                  ϕ                  1                )                            −        m                  ω                  2                          y                  0                cos        ⁡        (        ω        t        +                  ϕ                  2                )                            =        −        k        (                  x                  0                cos        ⁡        (        ω        t        +                  ϕ                  1                )        −                  y                  0                cos        ⁡        (        ω        t        +                  ϕ                  2                )        )        −                              m            g                    ℓ                cos        ⁡        (        ω        t        +                  ϕ                  2                )            However, without assuming that       ϕ      1    =      ϕ      2  , in which case everything factors out nicely to leave a simultaneous equation in       x      0   and       y      0  , I cannot see a way of making it linear in       x      0   and       y      0  . So am I expected to use this assumption or is there a mathematical way of simplifying it?If it is the former, then what would the physical justification for this assumption be?

Gary Salinas · Accepted Answer

If you add the two equations the terms in   k disappear and the terms in   cos distribute. You are left for an equation in   x  +  y. Then subtract the two equations and again the cosine terms distribute, leaving an equation in   x  −  y with a different frequency. You never need to assume       ϕ    1    =      ϕ    2   In fact, you can set one to zero, which just says you start your timer when that pendulum is at angle zero. You will eventually discover that the other is zero as well.

wo identitical pendula each of length l and with bobs of mass m are free to oscillate in the

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