| ### 1. Identification of Normal Modes | | ### 1. Identification of Normal Modes |
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| The given equations describing the positions of the two coupled pendulums are: | | The given equations describing the positions of the two coupled pendulums are: |
| $$x_1(t) = B \cos (\omega_0 t + \phi) + A \cos \omega t$$ | | $$x_1(t) = B \cos (\omega_0 t + \phi) + A \cos \omega t$$ |
| $$x_2(t) = B \cos (\omega_0 t + \phi) - A \cos \omega t$$ | | $$x_2(t) = B \cos (\omega_0 t + \phi) - A \cos \omega t$$ |
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| These equations show that the motion is a linear combination (superposition) of two distinct normal modes: | | These equations show that the motion is a linear combination (superposition) of two distinct normal modes: |
| 1. <b>Symmetric Mode ($\omega_0$):</b> When $A = 0$, both pendulums move in perfect phase with identical displacements ($x_1 = x_2$). In this mode, the connecting spring undergoes no stretching or compression, meaning it exerts zero force. The frequency is purely dictated by gravity and the length $l$ of the pendulums: | | 1. <b>Symmetric Mode ($\omega_0$):</b> When $A = 0$, both pendulums move in perfect phase with identical displacements ($x_1 = x_2$). In this mode, the connecting spring undergoes no stretching or compression, meaning it exerts zero force. The frequency is purely dictated by gravity and the length $l$ of the pendulums: |
| $$\omega_0^2 = \frac{g}{l}$$ | | $$\omega_0^2 = \frac{g}{l}$$ |
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| 2. <b>Antisymmetric Mode ($\omega$):</b> When $B = 0$, the pendulums move completely out of phase with opposite displacements ($x_1 = -x_2$). Here, the spring is actively stretched and compressed, providing an additional restoring force that increases the oscillation frequency ($\omega > \omega_0$). | | 2. <b>Antisymmetric Mode ($\omega$):</b> When $B = 0$, the pendulums move completely out of phase with opposite displacements ($x_1 = -x_2$). Here, the spring is actively stretched and compressed, providing an additional restoring force that increases the oscillation frequency ($\omega > \omega_0$). |
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| ### 2. Equations of Motion for Small Oscillations | | ### 2. Equations of Motion for Small Oscillations |
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| Let $m$ be the mass of each pendulum bob, $l$ be the length of the suspension strings, and $k$ be the stiffness of the coupling spring. For small angular deflections, the restoring force due to gravity is approximately $-mg\frac{x}{l} = -m\omega_0^2 x$. The force exerted by the spring on the first mass is $-k(x_1 - x_2)$ and on the second mass is $-k(x_2 - x_1)$. | | Let $m$ be the mass of each pendulum bob, $l$ be the length of the suspension strings, and $k$ be the stiffness of the coupling spring. For small angular deflections, the restoring force due to gravity is approximately $-mg\frac{x}{l} = -m\omega_0^2 x$. The force exerted by the spring on the first mass is $-k(x_1 - x_2)$ and on the second mass is $-k(x_2 - x_1)$. |
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| Setting up the equations of motion for both masses: | | Setting up the equations of motion for both masses: |
| $$m \ddot{x}_1 = -m\omega_0^2 x_1 - k(x_1 - x_2)$$ | | $$m \ddot{x}_1 = -m\omega_0^2 x_1 - k(x_1 - x_2)$$ |
| $$m \ddot{x}_2 = -m\omega_0^2 x_2 - k(x_2 - x_1)$$ | | $$m \ddot{x}_2 = -m\omega_0^2 x_2 - k(x_2 - x_1)$$ |
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| ### 3. Deriving Stiffness from the Antisymmetric Channel | | ### 3. Deriving Stiffness from the Antisymmetric Channel |
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| To isolate the antisymmetric parameters, we subtract the second equation of motion from the first: | | To isolate the antisymmetric parameters, we subtract the second equation of motion from the first: |
| $$m(\ddot{x}_1 - \ddot{x}_2) = -m\omega_0^2(x_1 - x_2) - 2k(x_1 - x_2)$$ | | $$m(\ddot{x}_1 - \ddot{x}_2) = -m\omega_0^2(x_1 - x_2) - 2k(x_1 - x_2)$$ |
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| Let us define the coordinate for relative displacement as $u = x_1 - x_2$. Dividing the entire relation by $m$ gives: | | Let us define the coordinate for relative displacement as $u = x_1 - x_2$. Dividing the entire relation by $m$ gives: |
| $$\ddot{u} = -\left(\omega_0^2 + \frac{2k}{m}\right)u \implies \ddot{u} + \left(\omega_0^2 + \frac{2k}{m}\right)u = 0$$ | | $$\ddot{u} = -\left(\omega_0^2 + \frac{2k}{m}\right)u \implies \ddot{u} + \left(\omega_0^2 + \frac{2k}{m}\right)u = 0$$ |
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| This represents a simple harmonic equation governing the antisymmetric mode. The square of its natural angular frequency must match the coefficient of $u$: | | This represents a simple harmonic equation governing the antisymmetric mode. The square of its natural angular frequency must match the coefficient of $u$: |
| $$\omega^2 = \omega_0^2 + \frac{2k}{m}$$ | | $$\omega^2 = \omega_0^2 + \frac{2k}{m}$$ |
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| ### 4. Solving for Spring Stiffness $k$ | | ### 4. Solving for Spring Stiffness $k$ |
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| Isolating the term containing the unknown stiffness $k$ from the frequency equation: | | Isolating the term containing the unknown stiffness $k$ from the frequency equation: |
| $$\omega^2 - \omega_0^2 = \frac{2k}{m}$$ | | $$\omega^2 - \omega_0^2 = \frac{2k}{m}$$ |
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| Multiplying both sides by $m$ and dividing by $2$ directly yields the stiffness of the connecting spring: | | Multiplying both sides by $m$ and dividing by $2$ directly yields the stiffness of the connecting spring: |
| $$k = \frac{m(\omega^2 - \omega_0^2)}{2}$$ | | $$k = \frac{m(\omega^2 - \omega_0^2)}{2}$$ |
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| #### Answer | | #### Answer |
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| $$\boxed{k = \frac{m(\omega^2 - \omega_0^2)}{2}}$$ | | $$\boxed{k = \frac{m(\omega^2 - \omega_0^2)}{2}}$$ |
