Let's tackle part (a) in **two steps** with explanations after each, and finish with a final answer. --- ## **Step 1: Write Equations for Displacement, Velocity, and Acceleration** **Given:** - Forcing function: \( F = F_ \cos(\omega t) \) - Equation of motion: \[ F_ \cos(\omega t) = m\ddot{x} + c\dot{x} + kx \] **Let:** - Displacement: \( x(t) = X \cos(\omega t - \phi) \) Where \( X \) is the amplitude, and \( \phi \) is the phase lag. **Velocity:** \[ \dot{x}(t) = -X \omega \sin(\omega t - \phi) \] **Acceleration:** \[ \ddot{x}(t) = -X \omega^2 \cos(\omega t - \phi) \] **Explanation:** We assume a solution for \( x \) that lags behind the force by phase \( \phi \) (due to damping). Taking derivatives gives us velocity and acceleration. --- ## **Step 2: Express Equation Using Cosine Terms and Draw Vector Diagram** **Substitute** \( x, \dot{x}, \ddot{x} \) into the equation (using only cosine terms): - \( \cos(\omega t) = \cos(\omega t - \phi + \phi) \) - Use trigonometric identities to express all terms as cosines: \[ \begin{align*} m\ddot{x} + c\dot{x} + kx &= m[-X \omega^2 \cos(\omega t - \phi)] + c[-X \omega \sin(\omega t - \phi)] + k[X \cos(\omega t - \phi)] \\ &= -mX \omega^2 \cos(\omega t - \phi) - cX \omega \sin(\omega t - \phi) + kX \cos(\omega t - \phi) \end{align*} \] Express \(\sin(\omega t - \phi)\) as a cosine with a phase shift: \[ \sin(\omega t - \phi) = \cos(\omega t - \phi - 90^\circ) \] So, the terms can be represented as **vectors** (phasors): - **Spring force (\( kx \))**: along \( \cos(\omega t - \phi) \) - **Damping force (\( c\dot{x} \))**: along \( \sin(\omega t - \phi) \), 90° out of phase - **Inertia (\( m\ddot{x} \))**: along \( -\cos(\omega t - \phi) \) **Vector Diagram:** - Draw all three vectors (phasors) from a common origin: - Spring force: horizontal (reference) - Damping force: vertical (lags by 90°) - Inertia: horizontal, opposite direction (lags by 180°) **Explanation:** Each term in the motion equation corresponds to a force vector (phasor) with a specific direction relative to the displacement. The force from the spring is in phase with displacement, damping force lags by 90°, and inertia is 180° out of phase. --- ## **Final Answer for (a):** ### **Equations:** - \( x(t) = X \cos(\omega t - \phi) \) - \( \dot{x}(t) = -X \omega \sin(\omega t - \phi) \) - \( \ddot{x}(t) = -X \omega^2 \cos(\omega t - \phi) \) ### **Phasor (Vector) Diagram:** - Draw vectors for \( kX \) (in phase), \( cX\omega \) (90° lag), and \( mX\omega^2 \) (180° lag). - The sum of these vectors equals the applied force phasor \( F_ \) at angle \( \omega t \). --- *This answers part (a) stepwise, providing equations and how to express them as a vector diagram, with explanations for each step.*

Sure, let's continue solving the problem step-by-step, addressing all parts as requested. --- ## **Part (b): Rearranging the Vector Diagram** ### **Step 1: Vector Sum of Zero** To achieve a vector sum of zero, we can express the forces in terms of their magnitudes and angles: 1. **Spring Force:** \( F_s = kX \) at angle \( 0^\circ \) (in phase with displacement). 2. **Damping Force:** \( F_d = c\dot{x} = c(-X\omega \sin(\omega t - \phi)) \) at angle \( 90^\circ \) (lagging). 3. **Inertia Force:** \( F_i = m\ddot{x} = -mX\omega^2 \cos(\omega t - \phi) \) at angle \( 180^\circ \) (opposing displacement). The relationship can be expressed as: \[ F = F_s + F_d + F_i = 0 \] ### **Step 2: Professional Sketch** - **Draw the vectors**: - Spring force \( F_s \) horizontally to the right. - Damping force \( F_d \) vertically upwards. - Inertia force \( F_i \) horizontally to the left. **Explanation:** The balance of these forces at steady-state means that the system is in equilibrium at the natural frequency, where the maximum transfer of force to displacement occurs, leading to maximum amplitude \( X \). --- ## **Final Answer for (b):** ### **Vector Sum of Zero:** - \( F_s + F_d + F_i = 0 \) ### **Sketch:** - Illustrates the forces with correct angles: \( F_s \) at \( 0^\circ \), \( F_d \) at \( 90^\circ \), \( F_i \) at \( 180^\circ \). --- ## **Part (c): Mechanisms Limiting Amplitude** ### **Discussion of Other Mechanisms:** 1. **Nonlinear Stiffness:** - As displacement increases, the effective stiffness may change, leading to a nonlinear response that limits amplitude. 2. **Mass Variation:** - If the mass changes (due to added weight or material fatigue), it can influence the system's natural frequency and thus the amplitude. 3. **Geometric Constraints:** - Physical constraints in the system may limit how far the mass can move, regardless of damping effects. 4. **External Friction:** - Friction from bearings or environmental factors can dissipate energy, limiting amplitude. **Explanation:** These mechanisms can affect the system's response both above and below the natural frequency, potentially leading to unexpected behaviors in amplitude. --- ## **Part (d): Expressing Resonant Frequency** ### **Step 1: Formula Derivation** The resonant frequency (\(\omega_r\)) can be expressed as: \[ \omega_r = \omega_n \sqrt{1 - \zeta^2} \] where: - \(\omega_n = \sqrt{\frac{k}{m}}\) is the natural frequency. - \(\zeta = \frac{c}{2\sqrt{mk}}\) is the damping ratio. ### **Step 2: Final Expression** Thus, the resonant frequency is: \[ \omega_r = \sqrt{\frac{k}{m}} \sqrt{1 - \left(\frac{c}{2\sqrt{mk}}\right)^2} \] **Explanation:** The resonant frequency is reduced from the natural frequency due to damping effects, which can alter the system's response characteristics. --- ## **Final Answer for (d):** ### **Resonant Frequency:** \[ \omega_r = \omega_n \sqrt{1 - \zeta^2} \] --- ## **Part (e): Extracting \( X/F \) and Transmissibility Ratio** ### **Step 1: Expressing \( X/F \)** Using the vector diagram: \[ \frac{X}{F} = \frac{1}{\left( \frac{k}{m} - \omega^2 \right) + j\left( \frac{c}{m} \omega \right)} \] ### **Step 2: Transmissibility Ratio** The transmissibility ratio is given by: \[ \frac{F_y}{F_0} = \frac{X}{X_0} = \frac{1}{\sqrt{\left(1 - \left(\frac{\omega}{\omega_n}\right)^2\right)^2 + \left(2\zeta \frac{\omega}{\omega_n}\right)^2}} \] **Explanation:** This ratio indicates how much of the input force is transmitted to the output displacement, especially significant at resonant frequency. --- ## **Final Answer for (e):** ### **Transmissibility Ratio:** \[ \frac{F_y}{F_0} = \frac{X}{X_0} = \frac{1}{\sqrt{\left(1 - \left(\frac{\omega}{\omega_n}\right)^2\right)^2 + \left(2\zeta \frac{\omega}{\omega_n}\right)^2}} \] --- This completes the solutions for all parts of the question, providing detailed steps and explanations for each.