Let's break down the problem step by step: **Given:** - Mass \( m = 180 \) kg - Spring stiffness \( k = 48 \) N/mm = \( 48,000 \) N/m - Damping coefficient \( c = 0.6 \) N·s/mm = \( 600 \) N·s/m - Initial velocity \( v_0 = 0.260 \) m/s - Initial displacement \( x_0 = 0 \) - Time \( t = 0.8 \) s --- ### Step 1: Write the equation of motion The equation for a damped spring-mass system is: \[ m\ddot{x} + c\dot{x} + kx = 0 \] --- ### Step 2: Calculate parameters - **Natural Frequency (\(\omega_n\)):** \[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{48000}{180}} = \sqrt{266.67} \approx 16.33 \text{ rad/s} \] - **Damping Ratio (\(\zeta\)):** \[ \zeta = \frac{c}{2\sqrt{km}} = \frac{600}{2 \sqrt{48000 \times 180}} \] \[ \sqrt{48000 \times 180} = \sqrt{8,640,000} \approx 2,940.7 \] \[ \zeta = \frac{600}{2 \times 2940.7} = \frac{600}{5881.4} \approx 0.102 \] - **Damped Natural Frequency (\(\omega_d\)):** \[ \omega_d = \omega_n \sqrt{1 - \zeta^2} \approx 16.33 \times \sqrt{1 - (0.102)^2} \] \[ \approx 16.33 \times \sqrt{0.9896} \approx 16.33 \times 0.9948 \approx 16.25 \text{ rad/s} \] --- ### Step 3: Solution for velocity Since initial displacement \( x_0 = 0 \), and \( v_0 = 0.26 \) m/s: \[ x(t) = e^{-\zeta \omega_n t} \left[ x_0 \cos(\omega_d t) + \frac{v_0 + \zeta \omega_n x_0}{\omega_d} \sin(\omega_d t) \right] \] For velocity: \[ \dot{x}(t) = \frac{d}{dt} x(t) \] Since \( x_0 = 0 \): \[ x(t) = e^{-\zeta \omega_n t} \left( \frac{v_0}{\omega_d} \sin(\omega_d t) \right) \] Take the derivative: \[ \dot{x}(t) = \frac{d}{dt} \left[ e^{-\zeta \omega_n t} \frac{v_0}{\omega_d} \sin(\omega_d t) \right] \] Apply product rule: \[ \dot{x}(t) = e^{-\zeta \omega_n t} \frac{v_0}{\omega_d} \left[ \omega_d \cos(\omega_d t) - \zeta \omega_n \sin(\omega_d t) \right] \] \[ = v_0 e^{-\zeta \omega_n t} \left[ \cos(\omega_d t) - \frac{\zeta \omega_n}{\omega_d} \sin(\omega_d t) \right] \] --- ### Step 4: Plug in values \[ v_0 = 0.26 \text{ m/s} \] \[ \zeta = 0.102 \] \[ \omega_n = 16.33 \] \[ \omega_d = 16.25 \] \[ t = 0.8 \] First, calculate the exponential term: \[ e^{-\zeta \omega_n t} = e^{-0.102 \times 16.33 \times 0.8} = e^{-1.333} \approx 0.263 \] Calculate \( \omega_d t \): \[ \omega_d t = 16.25 \times 0.8 = 13.0 \text{ radians} \] Now, evaluate the trigonometric terms: \[ \cos(13.0) \approx 0.974 \] \[ \sin(13.0) \approx 0.225 \] Calculate \(\frac{\zeta \omega_n}{\omega_d}\): \[ \frac{0.102 \times 16.33}{16.25} = \frac{1.664}{16.25} \approx 0.102 \] Now plug in: \[ \dot{x}(0.8) = 0.26 \times 0.263 \left[ 0.974 - 0.102 \times 0.225 \right] \] \[ = 0.26 \times 0.263 \left[ 0.974 - 0.023 \right] \] \[ = 0.26 \times 0.263 \times 0.951 \] \[ = 0.26 \times 0.250 \approx 0.065 \text{ m/s} \] Convert to mm/s (\( \times 1000 \)): \[ 0.065 \text{ m/s} = 65.2 \text{ mm/s} \] --- ### **Final Answer** \[ \boxed{65.2 \text{ mm/s}} \] So, the correct option is **65.2 mm/s**.