# Step-by-Step Solution Let's solve the problem step by step, as requested: --- ## **Step 1: Understanding the Problem** - **A pendulum** of length \( L \) and mass \( m \) is suspended from a support. - The support moves with \( y) = Ysin(\omega t)). - There is viscous damping, with a damping coefficient \( C \). - **Goal:** Derive the equation of motion for \( \theta(t) \), then solve for \( \theta(t) \) under the assumption of small amplitudes. --- ## **Step 2: Free-Body Diagram and Kinematics** Let: - \(\theta\): Angle from the vertical - The **pivot moves as** \( y(t) = Y\sin(\omega t) \). The position of mass \( m \) (in inertial frame): - \( x_m = y(t) + L\sin\theta \) - \( y_m = -L\cos\theta \) --- ## **Step 3: Velocity of the Mass** Differentiate position with respect to time: \[ \dot{x}_m = \dot{y}(t) + L\cos\theta \cdot \dot{\theta} \] \[ \dot{y}_m = L\sin\theta \cdot \dot{\theta} \] --- ## **Step 4: Kinetic and Potential Energy** ### Kinetic Energy (T): \[ T = \frac{1}{2}m(\dot{x}_m^2 + \dot{y}_m^2) \] Plug in expressions: \[ T = \frac{1}{2}m \left[ (\dot{y}(t) + L\cos\theta \cdot \dot{\theta})^2 + (L\sin\theta \cdot \dot{\theta})^2 \right] \] \[ = \frac{1}{2}m \left[ \dot{y}^2 + 2L\dot{y}\cos\theta \dot{\theta} + L^2\dot{\theta}^2 (\cos^2\theta + \sin^2\theta) \right] \] \[ = \frac{1}{2}m \left[ \dot{y}^2 + 2L\dot{y}\cos\theta \dot{\theta} + L^2\dot{\theta}^2 \right] \] ### Potential Energy (V): \[ V = -mgL\cos\theta \] --- ## **Step 5: Lagrangian** \[ \mathcal{L} = T - V \] \[ \mathcal{L} = \frac{1}{2}m \left[ \dot{y}^2 + 2L\dot{y}\cos\theta \dot{\theta} + L^2\dot{\theta}^2 \right] + mgL\cos\theta \] --- ## **Step 6: Equation of Motion (Euler-Lagrange Equation)** \[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}}{\partial \theta} = Q_{\theta} \] Where \( Q_{\theta} \) is the generalized force (non-conservative), arising from damping: \[ Q_{\theta} = -C \dot{\theta} \] Find derivatives: ### \( \frac{\partial \mathcal{L}}{\partial \dot{\theta}} \): \[ = mL\dot{y}\cos\theta + mL^2\dot{\theta} \] ### \( \frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) \): \[ = mL\ddot{y}\cos\theta - mL\dot{y}\sin\theta\dot{\theta} + mL^2\ddot{\theta} \] ### \( \frac{\partial \mathcal{L}}{\partial \theta} \): \[ = -mL\dot{y}\sin\theta\dot{\theta} - mgL\\theta \] --- ## **Step 7: Plug into Euler-Lagrange** \[ mL\ddot{y}\cos\theta - mL\dot{y}\sin\theta\dot{\theta} + mL^2\ddot{\theta} - [-mL\dot{y}\sin\theta\dot{\theta} - mgL\sin\theta] = -C\dot{\theta} \] \[ mL\ddot{y}\cos\theta - mL\dot{y}\sin\theta\dot{\theta} + mL^2\ddot{\theta} + mL\dot{y}\sin\theta\dot{\theta} + mgL\sin\theta = -C\dot{\theta} \] \[ mL\ddot{y}\cos\theta + mL^2\ddot{\theta} + mgL\sin\theta = -C\dot{\theta} \] Divide both sides by \( mL^2 \): \[ \frac{\ddot{y}}{L}\cos\theta + \ddot{\theta} + \frac{g}{L}\sin\theta = -\frac{C}{mL^2}\dot{\theta} \] --- ## **Step 8: Small Angle Approximation** For small \( \theta \): - \( \sin\theta \approx \theta \) - \( \cos\theta \approx 1 \) \[ \ddot{\theta} + \frac{C}{mL^2}\dot{\theta} + \frac{g}{L}\theta = -\frac{\ddot{y}}{L} \] --- ## **Step 9: Substitute for \( y(t) \)** Given: \( y(t) = Y\sin(\omega t) \implies \ddot{y}(t) = -Y\omega^2\sin(\omega t) \) So, \[ \ddot{\theta} + \frac{C}{mL^2}\dot{\theta} + \frac{g}{L}\theta = \frac{Y\omega^2}{L}\sin(\omega t) \] --- ## **Step 10: Final Answer** ### **Equation of Motion:** \[ \boxed{ \ddot{\theta} + \frac{C}{mL^2}\dot{\theta} + \frac{g}{L}\theta = \frac{Y\omega^2}{L}\sin(\omega t) } \] --- ### **General Solution for \(\theta(t)\):** This is a standard, forced, damped harmonic oscillator equation. The solution is the sum of homogeneous and particular solutions: #### Homogeneous part: \[ \theta_h(t) = A e^{-\zeta \omega_ t} \cos(\omega_d t + \phi) \] where: - \( \omega_ = \sqrt{\frac{g}{L}} \) - \( 2\zeta\omega_ = \frac{C}{mL^2} \) - \( \omega_d = \omega_\sqrt{1 - \zeta^2} \) #### Particular (steady-state) part: Try \( \theta_p(t) = B\sin(\omega t) \) Plug into the equation to solve for \( B \): \[ -B\omega^2 + i\omega \frac{C}{mL^2}B + \frac{g}{L}B = \frac{Y\omega^2}{L} \] Magnitude of steady-state: \[ B = \frac{\frac{Y\omega^2}{L}}{\sqrt{(\frac{g}{L} - \omega^2)^2 + (\frac{C}{mL^2}\omega)^2}} \] --- ## **Summary** ### **Equation of Motion:** \[ \boxed{ \ddot{\theta} + \frac{C}{mL^2}\dot{\theta} + \frac{g}{L}\theta = \frac{Y\omega^2}{L}\sin(\omega t) } \] ### **Steady-State Solution:** \[ \boxed{ \theta_{ss}(t) = \frac{\frac{Y\omega^2}{L}}{\sqrt{\left(\frac{g}{L} - \omega^2\right)^2 + \left(\frac{C}{mL^2}\omega\right)^2}} \sin(\omega t + \phi) } \] where \( \phi \) is the phase lag. --- If you need the **full time-domain solution** or the value of \( \phi \), let me know!