Let's break down and solve each part step by step. --- # **Given Data** - Mass, \( m = 5\, \text{kg} \) - Damping coefficient, \( c = 75\, \text{Ns/m} \) - System is at **50% of critical damping** - **Unforced vibration** (\( P= \)) - **Downward acceleration given**: \( a = 2\, \text{m/s}^2 \) --- ## **1. Equation of Motion** The general equation for a damped mass-spring system: \[ m\ddot{x} + c\dot{x} + kx = \] --- ### **a. Critical Damping Coefficient** \[ c_{cr} = 2\sqrt{mk} \] Given \( c = .5\,c_{cr} \) (50% of critical damping): \[ .5c_{cr} = 75 \implies c_{cr} = 150\, \text{Ns/m} \] \[ 150 = 2\sqrt{5k} \] \[ \sqrt{5k} = 75 \] \[ 5k = 5625 \] \[ k = 1125\, \text{N/m} \] --- ### **b. Equation for \( x(t) \) (Underdamped System, \( \zeta = .5 \))** - Damping ratio: \(\zeta = \frac{c}{2\sqrt{mk}} = .5\) - Natural frequency: \(\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1125}{5}} = 15\, \text{rad/s}\) - Damped natural frequency: \(\omega_d = \omega_n\sqrt{1-\zeta^2} = 15\sqrt{1-.25} = 15 \times .866 = 12.99\, \text{rad/s}\) **General Solution:** \[ x(t) = e^{-\zeta \omega_n t} \left[ A\cos(\omega_d t) + B\sin(\omega_d t) \right] \] \[ x(t) = e^{-.5 \times 15 t} \left[ A\cos(12.99 t) + B\sin(12.99 t) \right] \] \[ x(t) = e^{-7.5 t} \left[ A\cos(12.99 t) + B\sin(12.99 t) \right] \] --- ### **c. Initial Conditions after Downward Acceleration** - **Initial displacement, \( x() = \)** (from static position) - **Initial velocity** caused by \( a = 2\, \text{m/s}^2 \): The force on the mass is \( F = m a = 5 \times 2 = 10\, \text{N} \). This force gives an initial velocity: \[ F = m \frac{dv}{dt} \] But if this force is applied instantaneously, it imparts an initial velocity \( v_ \). Since it's a downward acceleration from rest, the initial velocity is: \[ v_ = a \cdot \Delta t \] If the acceleration is applied instantaneously, then \( v_ = \). If applied for a short interval, more information is needed. **Assume it's an initial velocity downwards:** Let's denote the initial velocity as \( v_ \). But typically, if it is "given a downward acceleration of \( 2\, \text{m/s}^2 \) from rest," the immediate effect is an initial velocity: \[ v_ = 2\, \text{m/s} \] (Assume for 1 second). --- ### **d. Initial Conditions** \[ x() = \] \[ \dot{x}() = v_ = 2\, \text{m/s} \] --- ### **e. Solve for Constants \(A\), \(B\)** At \( t= \): \[ x() = A = \implies A= \] Velocity: \[ \dot{x}(t) = \frac{d}{dt}\left[ e^{-7.5 t} B\sin(12.99 t) \right] \] \[ = e^{-7.5 t} \left[ -7.5 B\sin(12.99 t) + B \cdot 12.99 \cos(12.99 t) \right] \] At \( t= \): \[ \dot{x}() = -7.5 B \cdot + B \cdot 12.99 \cdot 1 = 12.99 B \] \[ 2 = 12.99 B \implies B = \frac{2}{12.99} = .154 \] --- ### **f. Final Solution** \[ \boxed{ x(t) = .154\, e^{-7.5 t} \sin(12.99 t) } \] --- ## **2. Maximum Distances Upwards and Downwards** The amplitude of an underdamped oscillator decays with \( e^{-\zeta \omega_n t} \). The maximum displacement occurs at the first peak (downwards), then at the next peak (upwards), etc. The **first peak** (downwards) occurs when: \[ \sin(12.99 t) = 1 \implies 12.99 t_1 = \frac{\pi}{2} \implies t_1 = \frac{\pi}{2 \times 12.99} = .121\, \text{s} \] Maximum downward distance: \[ x_{\text{max, down}} = .154\, e^{-7.5 \times .121} \cdot 1 = .154\, e^{-.907} = .154 \times .404 = .0622\, \text{m} \] The next peak (upwards) is at \( \sin(12.99 t_2) = -1 \): \[ 12.99 t_2 = \frac{3\pi}{2} \implies t_2 = \frac{3\pi}{2 \times 12.99} = .363\, \text{s} \] Maximum upward distance: \[ x_{\text{max, up}} = .154\, e^{-7.5 \times .363} \cdot (-1) = -.154\, e^{-2.722} = -.154 \times .066 = -.0102\, \text{m} \] --- ## **Summary** ### **Equation of Motion:** \[ \boxed{ x(t) = .154\, e^{-7.5 t} \sin(12.99 t) } \] ### **Maximum Displacements:** - **Maximum Downward:** \( \boxed{.0622\, \text{m}} \) - **Maximum Upward:** \( \boxed{-.0102\, \text{m}} \) --- **Let me know if you need a plot or further details!**