# Problem Restatement **Given:** - A building modeled as a single degree of freedom (SDOF) spring-mass system. - Mass of building, \( m = 10^5 \) kg - Stiffness of each wall, \( k = 3.519 \times 10^6 \) N/m (total stiffness = \( 2k \)) - Ground motion: amplitude = .1 m, frequency = 7.5 rad/s **Find:** The magnitude of the deflection of the top of the building. --- ## **Step 1: Model the System** ### Equation of Motion For a mass-spring system with base excitation: \[ m\ddot{x} + c\dot{x} + kx = -m\ddot{y}_g \] where \( y_g \) is ground displacement. Given there is **no mention of damping**, set \( c = \): \[ m\ddot{x} + kx = -m\ddot{y}_g \] Assume ground motion is harmonic: \[ y_g(t) = Y \sin(\omega t) \] where: - \( Y = .1 \) m (amplitude) - \( \omega = 7.5 \) rad/s So, \[ \ddot{y}_g = -Y \omega^2 \sin(\omega t) \] Plug into the equation: \[ m\ddot{x} + kx = mY\omega^2 \sin(\omega t) \] The steady-state solution (particular solution) is: \[ x(t) = X \sin(\omega t) \] Plug into the equation, solve for \( X \): \[ -m\omega^2 X \sin(\omega t) + k X \sin(\omega t) = m Y \omega^2 \sin(\omega t) \] \[ X (k - m\omega^2) = m Y \omega^2 \] \[ X = \frac{m Y \omega^2}{k - m\omega^2} \] ### **Total Stiffness** There are two walls, so: \[ k_{\text{total}} = 2 \times 3.519 \times 10^6 = 7.038 \times 10^6~\text{N/m} \] --- ## **Step 2: Plug in the Numbers** \[ X = \frac{m Y \omega^2}{k - m\omega^2} \] Given: - \( m = 10^5~\text{kg} \) - \( Y = .1~\text{m} \) - \( \omega = 7.5~\text{rad/s} \) - \( k = 7.038 \times 10^6~\text{N/m} \) Calculate \( m\omega^2 \): \[ m\omega^2 = 10^5 \times (7.5)^2 = 10^5 \times 56.25 = 5.625 \times 10^6 \] Calculate denominator: \[ k - m\omega^2 = 7.038 \times 10^6 - 5.625 \times 10^6 = 1.413 \times 10^6 \] Calculate numerator: \[ mY\omega^2 = 10^5 \times .1 \times (7.5)^2 = 10^5 \times .1 \times 56.25 = 10^4 \times 56.25 = 5.625 \times 10^5 \] \[ X = \frac{5.625 \times 10^5}{1.413 \times 10^6} \approx .398~\text{m} \] --- ## **Step 3: Graphical Representation** ### Frequency Response Plot Below is a plot showing the amplitude ratio versus driving frequency (normalized by natural frequency): ```plaintext |\ | \ | \ | \ | \ |-----\------------------- ``` - The resonance would occur as \( \omega \to \omega_n \). - In this case, \( \omega \) is less than \( \omega_n \) (see below). ### **Natural Frequency** \[ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{7.038 \times 10^6}{10^5}} = \sqrt{70.38} \approx 8.4~\text{rad/s} \] So, \( \omega = 7.5~\text{rad/s} < \omega_n \). --- ## **Final Answer** ### **Magnitude of the Deflection:** \[ \boxed{.398~\text{m}} \] --- ## **Summary Table** | Parameter | Value | |----------------|-------------------| | Mass \(m\) | \(10^5\) kg | | Stiffness \(k\)| \(7.038 \times 10^6\) N/m | | Ground Amplitude \(Y\) | .1 m | | Frequency \(\omega\) | 7.5 rad/s | | Deflection \(X\) | .398 m | --- ## **Step-by-Step Recap with Example** 1. **Write the forced vibration equation and solve for steady-state amplitude, using given parameters.** 2. **Plug in the values and compute the final deflection.** --- ### **Graph (Example)**  *Amplitude ratio vs normalized frequency (\(\omega/\omega_n\)); operating point at \(\omega/\omega_n \approx .89\).* --- **Final Deflection of the Building Top:** \[ \boxed{.398~\text{m}} \]