$3.4.18.$ Small oscillations of pendulums connected by a spring occur according to the law: $x_1 = B \cos (\omega_0 t + \phi) + A \cos\omega t$, $x_2 = B \cos (\omega_0 t + \phi) - A \cos\omega t$. Determine the stiffness of the spring linking the pendulums. In the equilibrium position the pendulums are vertical, the mass of each ball is $m$.
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### Solution
### 1. Identification of Normal Modes
The given equations describing the positions of the two coupled pendulums are:
$$x_1(t) = B \cos (\omega_0 t + \phi) + A \cos\omega t$$
@@ -12,18 +13,16 @@1. Identification of Normal Modes
$$x_2(t) = B \cos (\omega_0 t + \phi) - A \cos\omega t$$
These equations show that the motion is a linear combination (superposition) of two distinct normal modes:
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1.**Symmetric Mode ($\omega_0$):** 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:
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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:
$$\omega_0^2 = \frac{g}{l}$$
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2.**Antisymmetric Mode ($\omega$):** 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.<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
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:
This represents a simple harmonic equation governing the antisymmetric mode. The square of its natural angular frequency must match the coefficient of $u$:
$$\omega^2 = \omega_0^2 + \frac{2k}{m}$$
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### 4. Solving for Spring Stiffness $k$
Isolating the term containing the unknown stiffness $k$ from the frequency equation:
$$\omega^2 - \omega_0^2 = \frac{2k}{m}$$
Multiplying both sides by $m$ and dividing by $2$ directly yields the stiffness of the connecting spring:
$3.4.18.$ Small oscillations of pendulums connected by a spring occur according to the law: $x_1 = B \cos (\omega_0 t + \phi) + A \cos\omega t$, $x_2 = B \cos (\omega_0 t + \phi) - A \cos\omega t$. Determine the stiffness of the spring linking the pendulums. In the equilibrium position the pendulums are vertical, the mass of each ball is $m$.
$3.4.18.$ Small oscillations of pendulums connected by a spring occur according to the law: $x_1 = B \cos (\omega_0 t + \phi) + A \cos\omega t$, $x_2 = B \cos (\omega_0 t + \phi) - A \cos\omega t$. Determine the stiffness of the spring linking the pendulums. In the equilibrium position the pendulums are vertical, the mass of each ball is $m$.
### Solution
### Solution
### 1. Identification of Normal Modes
### 1. Identification of Normal Modes
The given equations describing the positions of the two coupled pendulums are:
The given equations describing the positions of the two coupled pendulums are:
$$x_1(t) = B \cos (\omega_0 t + \phi) + A \cos\omega t$$
$$x_1(t) = B \cos (\omega_0 t + \phi) + A \cos\omega t$$
@@ -12,18 +13,16 @@1. Identification of Normal Modes
$$x_2(t) = B \cos (\omega_0 t + \phi) - A \cos\omega t$$
$$x_2(t) = B \cos (\omega_0 t + \phi) - A \cos\omega t$$
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.**Symmetric Mode ($\omega_0$):** 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:
$$\omega_0^2 = \frac{g}{l}$$
$$\omega_0^2 = \frac{g}{l}$$
2.**Antisymmetric Mode ($\omega$):** 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
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 $-k(x_2 - x_1)$.
Setting up the equations of motion for both masses:
Setting up the equations of motion for both masses:
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 $k$
Isolating the term containing the unknown stiffness $k$ from the frequency equation:
Isolating the term containing the unknown stiffness $k$ from the frequency equation:
$$\omega^2 - \omega_0^2 = \frac{2k}{m}$$
$$\omega^2 - \omega_0^2 = \frac{2k}{m}$$
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:
