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Velocity transfer function of two-mass system with analogies

Short answer

Solve equations for masses, X2/F = (m1s^2+B1s+k1)/Δ with 4th-order Δ. Velocity output: V2(s)/F(s) = s(m1s^2+B1s+k1)/Δ. F-V analogy: e for F, L for m, R for B, C=1/k; k2 & k3 act as series capacitors 1/(k2+k3). F-I analogy uses current source, C for m, 1/L for k; output V2 is node 2 voltage.

Find the transfer function of mechanical system given below taking output as velocity of mass M2; also draw F-V and F-I analogy electrical networks. [Figure: left wall – damper B1 – mass m1 (displacement x1) – spring k1 – mass m2 (displacement x2, force f(t) applied); m2 also connected to the left wall through spring k2; m2 connected to the right wall through damper B2 and spring k3 in parallel]

Displacements x1x_1 (m1m_1) and x2x_2 (m2m_2, force ff). Output: velocity of m2m_2, v2=x˙2v_2 = \dot x_2, so V2(s)=sX2(s)V_2(s) = sX_2(s).

Equations of motion

  • m1m_1: damper B1B_1 to the left wall, spring k1k_1 to m2m_2.
  • m2m_2: spring k1k_1 to m1m_1, k2k_2 to the left wall, B2∥k3B_2 \parallel k_3 to the right wall.
m1x¨1+B1x˙1+k1(x1−x2)=0m2x¨2+B2x˙2+(k2+k3)x2+k1(x2−x1)=f(t)\begin{aligned} m_1\ddot x_1 + B_1\dot x_1 + k_1(x_1 - x_2) &= 0 \\ m_2\ddot x_2 + B_2\dot x_2 + (k_2 + k_3)x_2 + k_1(x_2 - x_1) &= f(t) \end{aligned}

Laplace form:

(m1s2+B1s+k1)X1−k1X2=0−k1X1+(m2s2+B2s+k1+k2+k3)X2=F\begin{aligned} (m_1s^2 + B_1s + k_1)X_1 - k_1X_2 &= 0 \\ -k_1X_1 + (m_2s^2 + B_2s + k_1 + k_2 + k_3)X_2 &= F \end{aligned}

Transfer function

X2F=m1s2+B1s+k1Δ,Δ=(m1s2+B1s+k1)(m2s2+B2s+k1+k2+k3)−k12\frac{X_2}{F} = \frac{m_1s^2 + B_1s + k_1}{\Delta}, \qquad \Delta = (m_1s^2 + B_1s + k_1)(m_2s^2 + B_2s + k_1 + k_2 + k_3) - k_1^2

Velocity output:

V2(s)F(s)=s(m1s2+B1s+k1)Δ\frac{V_2(s)}{F(s)} = \frac{s(m_1s^2 + B_1s + k_1)}{\Delta}

with

Δ=  m1m2s4+(B1m2+B2m1)s3+[B1B2+m1(k1+k2+k3)+m2k1]s2+[B1(k1+k2+k3)+B2k1]s+k1(k2+k3)\begin{aligned} \Delta =\; & m_1m_2s^4 + (B_1m_2 + B_2m_1)s^3 + \left[B_1B_2 + m_1(k_1 + k_2 + k_3) + m_2k_1\right]s^2 \\ & + \left[B_1(k_1 + k_2 + k_3) + B_2k_1\right]s + k_1(k_2 + k_3) \end{aligned}

F-V analogy (mesh, velocity →\to current)

   L1=m1  R1=B1             L2=m2   R2=B2   C=1/(k2+k3)
 +-LLLL---/\/\--+----------LLLL----/\/\------||-----+
 |              |                                   |
 |           C1=1/k1                               (~) e
 |     i1       |              i2                   |
 +--------------+-----------------------------------+

(k2k_2 and k3k_3 both act on x2x_2 alone, so they combine into one capacitor 1/(k2+k3)1/(k_2+k_3), i.e. C2=1/k2C_2 = 1/k_2 and C3=1/k3C_3 = 1/k_3 in series.) The required output V2V_2 corresponds to mesh current i2i_2.

F-I analogy (node, velocity →\to voltage)

 v1      L1    v2
 +----+--LLLL--+----+----+----+----+
 |    |        |    |    |    |    |
 C1   R1       C2   R2   L2   L3  (i)
 |    |        |    |    |    |    |
 +----+--------+----+----+----+----+ gnd

(C1=m1C_1 = m_1, R1=1/B1R_1 = 1/B_1, L1=1/k1L_1 = 1/k_1, C2=m2C_2 = m_2, R2=1/B2R_2 = 1/B_2, L2=1/k2L_2 = 1/k_2, L3=1/k3L_3 = 1/k_3.) The output velocity v2v_2 is the voltage of node 2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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