Let's analyze the problem step by step: ### **Step 1: Understanding the System** We have two masses \(M_1\) and \(M_2\) connected by a **damper** (B) and a **spring** (K). An external force \(f(t)\) is applied to \(M_1\). We are asked for the **transfer function** \( \frac{X_2(s)}{F(s)} \). Given: - \( M_1 = 3 \) kg - \( M_2 = 13 \) kg - \( B = 16 \) Ns/m - \( K = 6 \) N/m ### **Step 2: Writing the Equations of Motion** Let \(x_1(t)\) and \(x_2(t)\) be the displacements of \(M_1\) and \(M_2\), respectively. #### **For \(M_1\):** \[ M_1 \ddot{x}_1 = f(t) - B(\dot{x}_1 - \dot{x}_2) - K(x_1 - x_2) \] #### **For \(M_2\):** \[ M_2 \ddot{x}_2 = B(\dot{x}_1 - \dot{x}_2) + K(x_1 - x_2) \] ### **Step 3: Laplace Transform (Zero Initial Conditions)** Replace derivatives: - \(\mathcal{L}\{\dot{x}(t)\} = sX(s)\) - \(\mathcal{L}\{\ddot{x}(t)\} = s^2 X(s)\) #### **For \(M_1\):** \[ M_1 s^2 X_1(s) = F(s) - B[sX_1(s) - sX_2(s)] - K[X_1(s) - X_2(s)] \] #### **For \(M_2\):** \[ M_2 s^2 X_2(s) = B[sX_1(s) - sX_2(s)] + K[X_1(s) - X_2(s)] \] ### **Step 4: Arrange Equations** #### \(M_1\): \[ M_1 s^2 X_1 + B s (X_2 - X_1) + K(X_2 - X_1) = F(s) \] \[ (M_1 s^2 + B s + K) X_1 - (B s + K) X_2 = F(s) \] #### \(M_2\): \[ M_2 s^2 X_2 - B s (X_1 - X_2) - K (X_1 - X_2) = 0 \] \[ - (B s + K) X_1 + (M_2 s^2 + B s + K) X_2 = 0 \] ### **Step 5: Write in Matrix Form** \[ \begin{bmatrix} M_1 s^2 + B s + K & - (B s + K) \\ - (B s + K) & M_2 s^2 + B s + K \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \end{bmatrix} = \begin{bmatrix} F(s) \\ 0 \end{bmatrix} \] ### **Step 6: Solve for \( X_2(s)/F(s) \)** From Cramer's rule: \[ \frac{X_2(s)}{F(s)} = \frac{ \begin{vmatrix} M_1 s^2 + B s + K & 1 \\ - (B s + K) & 0 \end{vmatrix} }{ \begin{vmatrix} M_1 s^2 + B s + K & - (B s + K) \\ - (B s + K) & M_2 s^2 + B s + K \end{vmatrix} } \] But it's easier to use substitution: From the 2nd equation: \[ - (B s + K) X_1 + (M_2 s^2 + B s + K) X_2 = 0 \implies X_1 = \frac{(M_2 s^2 + B s + K)}{(B s + K)} X_2 \] Substitute into the 1st equation: \[ (M_1 s^2 + B s + K) X_1 - (B s + K) X_2 = F(s) \] \[ (M_1 s^2 + B s + K) \left( \frac{M_2 s^2 + B s + K}{B s + K} X_2 \right) - (B s + K) X_2 = F(s) \] \[ \frac{(M_1 s^2 + B s + K)(M_2 s^2 + B s + K) - (B s + K)^2}{B s + K} X_2 = F(s) \] \[ X_2(s) = \frac{B s + K}{(M_1 s^2 + B s + K)(M_2 s^2 + B s + K) - (B s + K)^2} F(s) \] ### **Step 7: Substitute Values** - \(M_1 = 3\) - \(M_2 = 13\) - \(B = 16\) - \(K = 6\) So, \(Bs + K = 16s + 6\) Now: \[ (M_1 s^2 + B s + K)(M_2 s^2 + B s + K) - (B s + K)^2 \] \[ = (3s^2 + 16s + 6)(13s^2 + 16s + 6) - (16s + 6)^2 \] Let's expand: - \(3s^2 \cdot 13s^2 = 39s^4\) - \(3s^2 \cdot 16s = 48s^3\) - \(3s^2 \cdot 6 = 18s^2\) - \(16s \cdot 13s^2 = 208s^3\) - \(16s \cdot 16s = 256s^2\) - \(16s \cdot 6 = 96s\) - \(6 \cdot 13s^2 = 78s^2\) - \(6 \cdot 16s = 96s\) - \(6 \cdot 6 = 36\) Add them up: - \(s^4\): **39** - \(s^3\): \(48 + 208 = 256\) - \(s^2\): \(18 + 256 + 78 = 352\) - \(s\): \(96 + 96 = 192\) - constant: \(36\) Now, subtract \((16s + 6)^2 = 256s^2 + 192s + 36\): So, the denominator: - \(s^4\): 39 - \(s^3\): 256 - \(s^2\): \(352 - 256 = 96\) - \(s\): \(192 - 192 = 0\) - constant: \(36 - 36 = 0\) So denominator: \(39s^4 + 256s^3 + 96s^2\) The numerator: \(16s + 6\) ### **Step 8: Write the Transfer Function** \[ \frac{X_2(s)}{F(s)} = \frac{16s + 6}{39s^4 + 256s^3 + 96s^2} \] ### **Step 9: Match to Options** Option **a**: \[ \frac{16,000s + 6,000}{39,000s^4 + 256,000s^3 + 96,000s^2} \] Which is the same as above but multiplied by 1,000 (which is just a scaling for clarity). **Correct answer:** \[ \boxed{a} \] --- ## **Summary** - **Correct option is (a).** - The transfer function is: \[ \frac{X_2(s)}{F(s)} = \frac{16s + 6}{39s^4 + 256s^3 + 96s^2} \] - In the format given in the options, **option (a) is correct**.