Damped Oscillation Analysis: Door Closer Design for a Heavy Soundproof Door

Problem Statement

Consider the motion of a heavy sliding door with an automatic closing mechanism (door closer) installed in a next-generation soundproof room.

The door can be treated as a point mass of mass $m$m, moving along a horizontal $x$x-axis. The fully closed position is the origin $x = 0$x=0; the opening direction is positive.

The door closer contains a spring (restoring force $-kx$−kx) and an oil damper (viscous damping force $-\gamma v$−γv, where $v = dx/dt$v=dx/dt). No other horizontal forces act on the door (rail friction is negligible).

At time $t = 0$t=0, the door is opened to position $x = x_0$x=x0​ ($x_0 > 0$x0​>0) and released from rest.

By changing the temperature in the lab, the oil viscosity (and therefore the damping coefficient $\gamma$γ) is set to three different values:

Case 1 (Winter — high viscosity): $\gamma = \gamma_1$γ=γ1​ The door moves in overdamped motion. Let $x_1\,[\text{m}]$x1​[m] be the door's position at $t = \ln 2\,\text{s}$t=ln2s.

Case 2 (Optimal temperature): $\gamma = \gamma_2$γ=γ2​ The door closes as quickly as possible without oscillation — critical damping. Let $x_2\,[\text{m}]$x2​[m] be the door's position at $t = 0.5\,\text{s}$t=0.5s. This is expressed as $x_2 = Ce^{-1}$x2​=Ce−1 for some constant $C$C (where $e$e is Euler's number).

Case 3 (Summer — low viscosity): $\gamma = \gamma_3$γ=γ3​ The door moves in underdamped oscillation. Let $t_3\,[\text{s}]$t3​[s] be the first time the door passes through $x = 0$x=0 (fully closed position) after release. The square of $t_3$t3​ is expressed as $t_3^2 = \frac{P}{Q}\pi^2$t32​=QP​π2 for coprime positive integers $P, Q$P,Q ($\pi$π is the ratio of a circle's circumference to its diameter).

Analyze the motion in each case by solving the differential equation, and find the required numerical values.

Constraints

• Door mass: $m = 100\,\mathrm{kg}$m=100kg
• Spring constant: $k = 400\,\mathrm{N/m}$k=400N/m
• Initial displacement: $x_0 = 0.96\,\mathrm{m}$x0​=0.96m
• Case 1 damping coefficient: $\gamma_1 = 500\,\mathrm{N \cdot s/m}$γ1​=500N⋅s/m
• Case 2 damping coefficient: $\gamma_2 = 400\,\mathrm{N \cdot s/m}$γ2​=400N⋅s/m
• Case 3 damping coefficient: $\gamma_3 = 200\,\mathrm{N \cdot s/m}$γ3​=200N⋅s/m

Input Format

From each case, compute $N_1$N1​, $N_2$N2​, $N_3$N3​ as follows, and give their sum $N = N_1 + N_2 + N_3$N=N1​+N2​+N3​ as a positive integer:

• $N_1 = x_1 \times 100$N1​=x1​×100
• $N_2 = C \times 100$N2​=C×100
• $N_3 = P + Q$N3​=P+Q