GSO003 Problem 9

Equation of Motion

By Newton's second law:

m\frac{d^2x}{dt^2} + \gamma\frac{dx}{dt} + kx = 0

Substituting $m = 100$ , $k = 400$ and dividing by 100:

\frac{d^2x}{dt^2} + \frac{\gamma}{100}\frac{dx}{dt} + 4x = 0

Initial conditions for all cases: $x(0) = 0.96$ , $\dot{x}(0) = 0$ .

Case 1: Overdamping ( $\gamma_1 = 500\,\mathrm{N \cdot s/m}$ )

\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + 4x = 0

Characteristic equation: $\lambda^2 + 5\lambda + 4 = (\lambda + 1)(\lambda + 4) = 0$ , giving $\lambda = -1,\,-4$ (two distinct real roots — overdamped).

General solution:

x(t) = A_1 e^{-t} + A_2 e^{-4t}

Applying initial conditions:

A_1 + A_2 = 0.96, \quad -A_1 - 4A_2 = 0

Solving: $A_2 = -0.32$ , $A_1 = 1.28$ .

x(t) = 1.28e^{-t} - 0.32e^{-4t}

At $t = \ln 2\,\text{s}$ (using $e^{-\ln 2} = \frac{1}{2}$ , $e^{-4\ln 2} = \frac{1}{16}$ ):

x_1 = 1.28 \times \frac{1}{2} - 0.32 \times \frac{1}{16} = 0.64 - 0.02 = 0.62\,\text{m}

N_1 = 0.62 \times 100 = 62

Case 2: Critical Damping ( $\gamma_2 = 400\,\mathrm{N \cdot s/m}$ )

\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 4x = 0

Characteristic equation: $(\lambda + 2)^2 = 0$ , repeated root $\lambda = -2$ — critical damping.

General solution:

x(t) = (B_1 + B_2 t)e^{-2t}

Applying initial conditions:

B_1 = 0.96, \quad B_2 - 2B_1 = 0 \implies B_2 = 1.92

x(t) = 0.96(1 + 2t)e^{-2t}

At $t = 0.5\,\text{s}$ :

x_2 = 0.96(1 + 1)e^{-1} = 1.92e^{-1}\,\text{m}

So $C = 1.92$ .

N_2 = 1.92 \times 100 = 192

Case 3: Underdamping ( $\gamma_3 = 200\,\mathrm{N \cdot s/m}$ )

\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 4x = 0

Characteristic equation: $\lambda^2 + 2\lambda + 4 = 0$ , roots $\lambda = -1 \pm \sqrt{3}\,i$ — underdamped.

General solution:

x(t) = e^{-t}(D_1\cos\sqrt{3}t + D_2\sin\sqrt{3}t)

Applying initial conditions:

D_1 = 0.96, \quad -D_1 + \sqrt{3}D_2 = 0 \implies D_2 = \frac{0.96}{\sqrt{3}}

x(t) = e^{-t}\left(0.96\cos\sqrt{3}t + \frac{0.96}{\sqrt{3}}\sin\sqrt{3}t\right)

The first zero crossing occurs when the bracketed term equals zero (since $e^{-t} > 0$ ):

\sqrt{3}\cos\sqrt{3}t_3 + \sin\sqrt{3}t_3 = 0 \implies \tan\sqrt{3}t_3 = -\sqrt{3}

The smallest positive solution is $\sqrt{3}t_3 = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ , so:

t_3 = \frac{2\pi}{3\sqrt{3}}

t_3^2 = \frac{4\pi^2}{27} = \frac{4}{27}\pi^2

Since $\gcd(4, 27) = 1$ : $P = 4$ , $Q = 27$ .

N_3 = 4 + 27 = 31

Final Answer

N = N_1 + N_2 + N_3 = 62 + 192 + 31 = 285

Answer: 285

GSO003 Problem 9

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

Problem Statement

Constraints

Input Format

Solution

Equation of Motion

Case 1: Overdamping ( $\gamma_1 = 500\,\mathrm{N \cdot s/m}$ )

Case 2: Critical Damping ( $\gamma_2 = 400\,\mathrm{N \cdot s/m}$ )

Case 3: Underdamping ( $\gamma_3 = 200\,\mathrm{N \cdot s/m}$ )

Final Answer

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

Problem Statement

Constraints

Input Format

Solution

Equation of Motion

Case 1: Overdamping (γ1=500 N⋅s/m\gamma_1 = 500\,\mathrm{N \cdot s/m}γ1​=500N⋅s/m)

Case 2: Critical Damping (γ2=400 N⋅s/m\gamma_2 = 400\,\mathrm{N \cdot s/m}γ2​=400N⋅s/m)

Case 3: Underdamping (γ3=200 N⋅s/m\gamma_3 = 200\,\mathrm{N \cdot s/m}γ3​=200N⋅s/m)

Final Answer

Case 1: Overdamping ( $\gamma_1 = 500\,\mathrm{N \cdot s/m}$ )

Case 2: Critical Damping ( $\gamma_2 = 400\,\mathrm{N \cdot s/m}$ )

Case 3: Underdamping ( $\gamma_3 = 200\,\mathrm{N \cdot s/m}$ )