QuestionDecember 15, 2025

#4) A mass of 4 kg stretches spring 40 cm. The mass is acted on by an external force of 180e^-2t+196N and moves in a medium that imparts a viscous force of 34 N when the speed of the mass is (1)/(3)m/sec Mass is set in motion from equilibrium position with an initial velocity of 1m/sec A) Formulate an initial value problem (IVP) B) Solve the IVP

#4) A mass of 4 kg stretches spring 40 cm. The mass is acted on by an external force of 180e^-2t+196N and moves in a medium that imparts a viscous force of 34 N when the speed of the mass is (1)/(3)m/sec Mass is set in motion from equilibrium position with an initial velocity of 1m/sec A) Formulate an initial value problem (IVP) B) Solve the IVP
#4) A mass of 4 kg stretches spring 40 cm. The mass is acted on by an external force of 180e^-2t+196N and moves in a medium that imparts a viscous force of 34 N when the speed of the mass is (1)/(3)m/sec Mass is set in motion from equilibrium position with an initial velocity of 1m/sec A) Formulate an initial value problem (IVP) B) Solve the IVP

Solution
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Answer

**IVP:** 4x'' + 102x' + 98x = 180e^{-2t} + 196, x(0) = 0, x'(0) = 1. ### **Solution:** x(t) = -\frac{6}{47} e^{-t} + \frac{6}{47} e^{-24.5 t} - 2 e^{-2t} + 2. Explanation 1. Find the spring constant k The spring stretches 40 cm = 0.4 m under a force of mg = 4 \cdot 9.8 = 39.2 N. By **Hooke's law** F = kx, we get k = \frac{39.2}{0.4} = 98 \ \text{N/m}. 2. Find the damping coefficient c The viscous force is 34 N when v = \frac{1}{3} m/s. By F = c v, we get c = \frac{34}{1/3} = 102 N·s/m. 3. Write the differential equation For mass-spring-damper with external force: **m x'' + c x' + k x = F(t)** Here: 4x'' + 102x' + 98x = 180e^{-2t} + 196. 4. Initial conditions Given: x(0) = 0 (equilibrium start), x'(0) = 1 m/s. 5. Formulate the IVP 4x'' + 102x' + 98x = 180e^{-2t} + 196, x(0) = 0, x'(0) = 1. 6. Solve the homogeneous equation 4r^2 + 102r + 98 = 0 → divide by 2: 2r^2 + 51r + 49 = 0. Roots: r = \frac{-51 \pm \sqrt{2601 - 392}}{4} = \frac{-51 \pm \sqrt{2209}}{4} = \frac{-51 \pm 47}{4}. So r_1 = -1, r_2 = -24.5. Homogeneous solution: x_h(t) = C_1 e^{-t} + C_2 e^{-24.5 t}. 7. Solve for particular solution For 180e^{-2t}: try Ae^{-2t}. Sub into 4x'' + 102x' + 98x: 4(4A e^{-2t}) + 102(-2A e^{-2t}) + 98(A e^{-2t}) = (16A - 204A + 98A)e^{-2t} = (-90A) e^{-2t}. Set -90A = 180 → A = -2. For constant 196: try B. Sub into equation: 98B = 196 → B = 2. Particular solution: x_p(t) = -2 e^{-2t} + 2. 8. General solution x(t) = C_1 e^{-t} + C_2 e^{-24.5 t} - 2 e^{-2t} + 2. 9. Apply initial conditions x(0) = C_1 + C_2 - 2 + 2 = C_1 + C_2 = 0 → C_2 = -C_1. x'(t) = -C_1 e^{-t} - 24.5 C_2 e^{-24.5 t} + 4 e^{-2t}. x'(0) = -C_1 - 24.5(-C_1) + 4 = 23.5C_1 + 4 = 1 → 23.5 C_1 = -3 → C_1 = -\frac{3}{23.5} = -\frac{6}{47}, C_2 = \frac{6}{47}.

Explanation

1. Find the spring constant $k$<br /> The spring stretches 40 cm = 0.4 m under a force of $mg = 4 \cdot 9.8 = 39.2$ N. <br />By **Hooke's law** $F = kx$, we get $k = \frac{39.2}{0.4} = 98 \ \text{N/m}$.<br /><br />2. Find the damping coefficient $c$<br /> The viscous force is $34$ N when $v = \frac{1}{3}$ m/s. <br />By $F = c v$, we get $c = \frac{34}{1/3} = 102$ N·s/m.<br /><br />3. Write the differential equation<br /> For mass-spring-damper with external force: <br />**$m x'' + c x' + k x = F(t)$** <br />Here: $4x'' + 102x' + 98x = 180e^{-2t} + 196$.<br /><br />4. Initial conditions<br /> Given: $x(0) = 0$ (equilibrium start), $x'(0) = 1$ m/s.<br /><br />5. Formulate the IVP<br /> $4x'' + 102x' + 98x = 180e^{-2t} + 196$, $x(0) = 0$, $x'(0) = 1$.<br /><br />6. Solve the homogeneous equation<br /> $4r^2 + 102r + 98 = 0$ → divide by 2: $2r^2 + 51r + 49 = 0$. <br />Roots: $r = \frac{-51 \pm \sqrt{2601 - 392}}{4} = \frac{-51 \pm \sqrt{2209}}{4} = \frac{-51 \pm 47}{4}$. <br />So $r_1 = -1$, $r_2 = -24.5$.<br /><br />Homogeneous solution: $x_h(t) = C_1 e^{-t} + C_2 e^{-24.5 t}$.<br /><br />7. Solve for particular solution<br /> For $180e^{-2t}$: try $Ae^{-2t}$. Sub into $4x'' + 102x' + 98x$: <br />$4(4A e^{-2t}) + 102(-2A e^{-2t}) + 98(A e^{-2t}) = (16A - 204A + 98A)e^{-2t} = (-90A) e^{-2t}$. <br />Set $-90A = 180$ → $A = -2$.<br /><br />For constant $196$: try $B$. Sub into equation: $98B = 196$ → $B = 2$.<br /><br />Particular solution: $x_p(t) = -2 e^{-2t} + 2$.<br /><br />8. General solution<br /> $x(t) = C_1 e^{-t} + C_2 e^{-24.5 t} - 2 e^{-2t} + 2$.<br /><br />9. Apply initial conditions<br /> $x(0) = C_1 + C_2 - 2 + 2 = C_1 + C_2 = 0$ → $C_2 = -C_1$.<br /><br />$x'(t) = -C_1 e^{-t} - 24.5 C_2 e^{-24.5 t} + 4 e^{-2t}$. <br />$x'(0) = -C_1 - 24.5(-C_1) + 4 = 23.5C_1 + 4 = 1$ → $23.5 C_1 = -3$ → $C_1 = -\frac{3}{23.5} = -\frac{6}{47}$, $C_2 = \frac{6}{47}$.
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