Questions: II. Forced Vibration: Base Excitation 1. A trailer of mass m=1000 kg is pulled at a constant speed of v=50 km / h over a bumpy road surface modeled as a sinusoidal base excitation with a wavelength of λ=5 m and an amplitude Y=50 mm. The vertical displacement of the road acts as a base excitation to the suspension system. The effective stiffness of the suspension is k=350 kN / m, and the damping ratio is given as ζ=0.5. (1) Draw a free body diagram of the body, m. (2) Derive an equation of motion using Newton's law of motion. (3) Determine the undamped natural frequency, ωn. (4) Determine the period of the base excitation, τ. (5) Determine the frequency of the base excitation, ω. (6) Determine the frequency ratio, r. (7) Determine the amplitude of the vehicle, X. The amplitude ratio can be found from: X/Y=[(1+(2 ζ r)^2)/((1-r^2)^2+(2 ζ r)^2)]^(1 / 2)

(1) Draw a free body diagram of the body, \( m \). Free Body Diagram

  ↑ x
  |
  m

----o---- | | | k | |-------| c | ↓ ----- y(t)

\(\boxed{\text{See the diagram above.}}\)

(2) Derive an equation of motion using Newton's law of motion. Applying Newton's second law

The equation of motion for the system can be written as: \( m\ddot{x} + c(\dot{x} - \dot{y}) + k(x - y) = 0 \) \(\boxed{m\ddot{x} + c(\dot{x} - \dot{y}) + k(x - y) = 0}\)

(3) Determine the undamped natural frequency, \( \omega_{n} \). Formula for undamped natural frequency

\( \omega_n = \sqrt{\frac{k}{m}} \) Calculation

\( \omega_n = \sqrt{\frac{350 \times 10^3 \text{ N/m}}{1000 \text{ kg}}} = \sqrt{350} \text{ rad/s} \approx 18.71 \text{ rad/s} \) \(\boxed{\omega_n \approx 18.71 \text{ rad/s}}\)

(4) Determine the period of the base excitation, \( \tau \). Relationship between wavelength, speed, and period

\( \tau = \frac{\lambda}{v} \) Calculation

\( v = 50 \text{ km/h} = \frac{50 \times 1000}{3600} \text{ m/s} \approx 13.89 \text{ m/s} \) \( \tau = \frac{5 \text{ m}}{13.89 \text{ m/s}} \approx 0.36 \text{ s} \) \(\boxed{\tau \approx 0.36 \text{ s}}\)

(5) Determine the frequency of the base excitation, \( \omega \). Relationship between period and frequency

\( \omega = \frac{2\pi}{\tau} \) Calculation

\( \omega = \frac{2\pi}{0.36 \text{ s}} \approx 17.45 \text{ rad/s} \) \(\boxed{\omega \approx 17.45 \text{ rad/s}}\)

(6) Determine the frequency ratio, \( r \). Definition of frequency ratio

\( r = \frac{\omega}{\omega_n} \) Calculation

\( r = \frac{17.45 \text{ rad/s}}{18.71 \text{ rad/s}} \approx 0.93 \) \(\boxed{r \approx 0.93}\)

(7) Determine the amplitude of the vehicle, \( X \). Given formula

\( \left|\frac{X}{Y}\right|=\left[\frac{1+(2 \zeta r)^{2}}{\left(1-r^{2}\right)^{2}+(2 \zeta r)^{2}}\right]^{1 / 2} \) Calculation

\( \left|\frac{X}{Y}\right|=\left[\frac{1+(2 \times 0.5 \times 0.93)^{2}}{\left(1-0.93^{2}\right)^{2}+(2 \times 0.5 \times 0.93)^{2}}\right]^{1 / 2} \approx \left[\frac{1+0.8649}{0.0180+0.8649}\right]^{1/2} \approx 1.38 \) \( X = 1.38 \times Y = 1.38 \times 50 \text{ mm} = 69 \text{ mm} \) \(\boxed{X \approx 69 \text{ mm}}\)

\( \text{Free Body Diagram: See the diagram above.} \) \( m\ddot{x} + c(\dot{x} - \dot{y}) + k(x - y) = 0 \) \( \omega_n \approx 18.71 \text{ rad/s} \) \( \tau \approx 0.36 \text{ s} \) \( \omega \approx 17.45 \text{ rad/s} \) \( r \approx 0.93 \) \( X \approx 69 \text{ mm} \)