Differential Equations Quiz: Damping Types

What this quiz covers

This quiz focuses on Damping Types, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $R > 6$
2. $R = 6$
3. $0 \le R < 3$
4. $0 \le R < 6$ (correct answer)

Explanation: Substituting the given values, the differential equation is $q'' + Rq' + 9q = 0$. The circuit exhibits oscillations if the system is underdamped. This occurs when the discriminant of the characteristic equation $r^2 + Rr + 9 = 0$ is negative. The discriminant is $\Delta = R^2 - 4(1)(9) = R^2 - 36$. For the system to be underdamped, we need $R^2 - 36 < 0$, which simplifies to $R^2 < 36$. Since the resistance $R$ must be non-negative, the range for $R$ is $0 \le R < 6$.

Question 2

1. 4
2. 8 (correct answer)
3. 16
4. 32

Explanation: The condition of returning to equilibrium in the fastest possible time without oscillation describes a critically damped system. A system governed by $ay'' + by' + cy = 0$ is critically damped when the discriminant of its characteristic equation is zero, i.e., $b^2 - 4ac = 0$. For the given equation, $a=1$, $b=b$, and $c=16$. Thus, we need $b^2 - 4(1)(16) = 0$, which gives $b^2 = 64$. Since $b$ must be positive, $b=8$. The initial conditions do not affect the value of $b$ that results in critical damping.

Question 3

1. Overdamped
2. Underdamped (correct answer)
3. Still critically damped
4. Unstable with growing oscillations

Explanation: Initially, the system is critically damped, so the discriminant of the characteristic equation is zero: $\gamma^2 - 4mk = 0$, which means $\gamma^2 = 4mk$. The mass is then doubled to $m_{new} = 2m$. The new discriminant is $\Delta_{new} = \gamma^2 - 4(m_{new})k = \gamma^2 - 4(2m)k = \gamma^2 - 8mk$. Substituting $\gamma^2 = 4mk$ into the new discriminant gives $\Delta_{new} = 4mk - 8mk = -4mk$. Since mass $m$ and spring constant $k$ are positive, $\Delta_{new}$ is negative. A negative discriminant corresponds to an underdamped system.

Question 4

1. $r_1 + r_2 < 0$
2. $r_1 r_2 > 0$
3. $r_1 < 0$ and $r_2 < 0$ (correct answer)
4. No additional condition is needed.

Explanation: Distinct real roots mean the system is overdamped, and its solution is $y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$. For the system to be stable and return to equilibrium ($y(t) \to 0$ as $t \to \infty$), both exponential terms must decay. This requires both exponents to be negative, so we must have $r_1 < 0$ and $r_2 < 0$. While options A and B are consequences of this condition (if both roots are negative, their sum is negative and their product is positive), they are not sufficient on their own. For example, $r_1=1, r_2=-3$ satisfies $r_1+r_2 < 0$, but the $e^t$ term would lead to instability. For the given equation with $a,b,c > 0$, the roots are always negative or have negative real parts, so $r_1<0$ and $r_2<0$ is always true for overdamped systems of this form, but the question asks for the condition on the roots themselves.

Question 5

1. Overdamped (correct answer)
2. Underdamped
3. Still critically damped
4. Undamped

Explanation: First, find the value of $k$ for which the system is critically damped. The discriminant is $b^2 - 4ac = 4^2 - 4(1)k = 16 - 4k$. For critical damping, the discriminant is zero, so $16 - 4k = 0$, which gives $k=4$. Next, the value of $k$ decreases by 25%. The new value is $k_{new} = k - 0.25k = 0.75k = 0.75(4) = 3$. Now, we classify the new system with $k_{new}=3$: $y'' + 4y' + 3y = 0$. The new discriminant is $4^2 - 4(1)(3) = 16 - 12 = 4$. Since the discriminant is positive, the new system is overdamped.

Question 6

1. $\zeta > 1$
2. $\zeta = 1$
3. $0 < \zeta < 1$ (correct answer)
4. $\zeta = 0$

Explanation: A system exhibits decaying oscillations if it is underdamped. The condition for an underdamped system is that the discriminant of the characteristic equation is negative: $\Delta = \gamma^2 - 4mk < 0$. This inequality can be written as $\gamma^2 < 4mk$. Since $\gamma$ and $\sqrt{mk}$ are positive, taking the square root gives $0 < \gamma < 2\sqrt{mk}$. The damping ratio is defined as $\zeta = \frac{\gamma}{2\sqrt{mk}}$. Dividing the inequality by $2\sqrt{mk}$ gives $\frac{0}{2\sqrt{mk}} < \frac{\gamma}{2\sqrt{mk}} < \frac{2\sqrt{mk}}{2\sqrt{mk}}$, which simplifies to $0 < \zeta < 1$.

Question 7

1. System I is overdamped and System II is critically damped; System I returns to equilibrium more slowly than System II. (correct answer)
2. System I is underdamped and oscillates, while System II does not oscillate.
3. Both systems are overdamped, but System I returns to equilibrium faster.
4. System I is critically damped and System II is overdamped; System I returns to equilibrium faster than System II.

Explanation: When you encounter second-order linear differential equations with constant coefficients, you need to analyze the characteristic equation to determine the type of damping and compare how quickly systems return to equilibrium. For both systems, substitute $y = e^{rt}$ to get the characteristic equations. System I: $r^2 + 10r + 9 = 0$ factors to $(r+1)(r+9) = 0$, giving real roots $r_1 = -1, r_2 = -9$. Since the roots are real and distinct, System I is overdamped with solution $y(t) = c_1e^{-t} + c_2e^{-9t}$. System II: $r^2 + 6r + 9 = 0$ factors to $(r+3)^2 = 0$, giving repeated root $r = -3$. This makes System II critically damped with solution $z(t) = (c_1 + c_2t)e^{-3t}$. Applying initial conditions $y(0) = z(0) = 1$ and $y'(0) = z'(0) = 0$, you get $y(t) = \frac{9}{8}e^{-t} - \frac{1}{8}e^{-9t}$ and $z(t) = (1 + 3t)e^{-3t}$. The overdamped system is dominated by its slower exponential term $e^{-t}$, while the critically damped system decays as $e^{-3t}$. Since $e^{-3t}$ decays faster than $e^{-t}$, System II returns to equilibrium more quickly. Answer A correctly identifies the damping types and relative speeds. B is wrong because System I doesn't oscillate (real roots). C incorrectly calls both overdamped and reverses the speed comparison. D incorrectly swaps the damping classifications. Strategy tip: Remember that critically damped systems return to equilibrium fastest, followed by overdamped, then underdamped systems.

Question 8

1. 0
2. $4e^{-1}$
3. The displacement increases without bound.
4. 4 (correct answer)

Explanation: When you encounter a second-order linear homogeneous differential equation like $y'' + by' + 25y = 0$, you're dealing with oscillatory motion that can be underdamped, critically damped, or overdamped depending on the coefficient $b$. For critical damping, the discriminant of the characteristic equation $r^2 + br + 25 = 0$ must equal zero. This means $b^2 - 4(1)(25) = 0$, so $b^2 = 100$ and $b = 10$ (since $b > 0$). The characteristic equation becomes $r^2 + 10r + 25 = (r + 5)^2 = 0$, giving a repeated root $r = -5$. The general solution for critical damping is $y(t) = (c_1 + c_2t)e^{-5t}$. Using initial conditions $y(0) = 4$ and $y'(0) = 0$ (starts from rest), you get $c_1 = 4$ and $c_2 = 20$, so $y(t) = (4 + 20t)e^{-5t}$. To find the maximum, set $y'(t) = 20e^{-5t} - 5(4 + 20t)e^{-5t} = e^{-5t}(20 - 20 - 100t) = -100te^{-5t} = 0$. This occurs at $t = 0$, where $y(0) = 4$. Answer D is correct: the maximum displacement is 4, occurring at the initial moment. Answer A is wrong because the system doesn't immediately go to zero displacement. Answer B represents a common miscalculation involving the exponential factor. Answer C is incorrect because critical damping prevents oscillation and ensures the system approaches equilibrium. Remember: in critically damped systems starting from rest, maximum displacement occurs at $t = 0$ since the system monotonically approaches equilibrium without overshooting.

Question 9

1. 7
2. 8 (correct answer)
3. 9
4. 64

Explanation: The system is not underdamped if it is critically damped or overdamped. This occurs when the discriminant of the characteristic equation $2r^2 + \gamma r + 8 = 0$ is greater than or equal to zero. The discriminant is $\Delta = b^2 - 4ac = \gamma^2 - 4(2)(8) = \gamma^2 - 64$. The transition from underdamped to critically damped occurs when $\Delta = 0$, which means $\gamma^2 - 64 = 0$. Since $\gamma > 0$, we find $\gamma = 8$. For $\gamma=8$, the system is critically damped. For $\gamma > 8$, it is overdamped. Therefore, the system is not underdamped for $\gamma \ge 8$. The smallest integer value of $\gamma$ that satisfies this condition is 8.

Question 10

1. The original system was critically damped with $k = 9$, and the damping coefficient is $\gamma = 6$
2. The original system was overdamped with $k < 9$, and increasing $k$ reduced the discriminant below zero (correct answer)
3. The original system was underdamped with $k > 9$, and the spring constant change altered the natural frequency
4. The original system was overdamped with $k > 9$, but the 20% increase was insufficient to reach critical damping

Explanation: For the characteristic equation $r^2 + 6r + k = 0$, the discriminant is $\Delta = 36 - 4k$. Overdamped: $\Delta > 0$ ($k < 9$), critically damped: $\Delta = 0$ ($k = 9$), underdamped: $\Delta < 0$ ($k > 9$). If increasing $k$ by 20% changes the system from overdamped to underdamped, the original $k$ must have been less than 9, and the new value $1.2k$ must be greater than 9. This means $7.5 < k < 9$.

Question 11

1. The system would become critically damped since reducing damping moves it toward the critical damping boundary
2. The system would remain underdamped but with increased oscillation frequency and slower decay rate (correct answer)
3. The system would become overdamped because the natural frequency would decrease below the damping frequency
4. The system would remain underdamped but with decreased oscillation frequency and faster decay rate

Explanation: From the given solution $x(t) = e^{-3t}(A\cos(4t) + B\sin(4t))$, we identify the current system as underdamped with damping parameter $\alpha = 3$ and oscillation frequency $\omega_d = 4$. The natural frequency is $\omega_0 = \sqrt{\alpha^2 + \omega_d^2} = \sqrt{9 + 16} = 5$. The current damping coefficient is b = 2m\alpha = 6m$. If damping is reduced by half, b_{new} = 3m$, so$\alpha_{new} = 1.5$. The new damped frequency is$\omega_{d,new} = \sqrt{\omega_0^2 - \alpha_{new}^2} = \sqrt{25 - 2.25} = \sqrt{22.75} \approx 4.77 > 4$. The system remains underdamped with higher oscillation frequency and slower decay (smaller$\alpha$$).

Question 12

1. The damping parameter must satisfy $\lambda = \omega_0$ to ensure the square root term equals zero
2. The damping parameter must satisfy $\lambda < \omega_0$ to ensure complex characteristic roots
3. The damping parameter must satisfy $\lambda > \omega_0$ to ensure real characteristic roots and avoid complex exponentials (correct answer)
4. The damping parameter must satisfy $\lambda \geq \frac{\omega_0}{2}$ to ensure physical realizability of the exponential terms

Explanation: The given solution form $y(t) = e^{-\lambda t}(c_1 e^{\sqrt{\lambda^2 - \omega_0^2}t} + c_2 e^{-\sqrt{\lambda^2 - \omega_0^2}t})$ corresponds to the overdamped case where both characteristic roots are real and distinct. For the expression $\sqrt{\lambda^2 - \omega_0^2}$ to be real, we need $\lambda^2 - \omega_0^2 \geq 0$, which means $\lambda \geq \omega_0$. Since we want distinct roots (overdamped), we need $\lambda > \omega_0$. If $\lambda = \omega_0$, we get critical damping with a repeated root. If $\lambda < \omega_0$, we get underdamped motion with complex roots and oscillatory solutions.

Question 13

1. System 1 (underdamped) reaches 5% faster than System 2 (underdamped) because it has higher damping ratio (correct answer)
2. System 2 (underdamped) reaches 5% faster than System 1 (underdamped) because it has lower effective damping
3. System 1 (underdamped) and System 2 (critically damped) reach 5% in the same time due to identical natural frequencies
4. System 2 (underdamped) reaches 5% faster than System 1 (underdamped) because its exponential decay rate is higher

Explanation: System 1: $\zeta_1 = 0.6 < 1$ (underdamped), $\alpha_1 = \zeta_1 \omega_0 = 0.6 \times 6 = 3.6$. System 2: From roots $r = -2 \pm 4i$, we have $\alpha_2 = 2$ and $\omega_{d2} = 4$. Since $\omega_0 = 6$, we get $\zeta_2 = \frac{\alpha_2}{\omega_0} = \frac{2}{6} = 0.33 < 1$ (underdamped). The envelope decay is governed by $e^{-\alpha t}$. To reach 5% amplitude: $e^{-\alpha t} = 0.05$, so $t = \frac{\ln(20)}{\alpha}$. For System 1: $t_1 = \frac{\ln(20)}{3.6} \approx 0.83$ s. For System 2: $t_2 = \frac{\ln(20)}{2} \approx 1.5$ s. System 1 reaches 5% faster due to higher damping ratio and faster exponential decay.

Question 14

1. The circuit becomes underdamped with oscillation frequency $\omega_d = 2$ rad/s and exponential decay factor $e^{-0.75t}$
2. The circuit becomes overdamped with two distinct real time constants $\tau_1 = 2$ s and $\tau_2 = 0.5$ s
3. The circuit remains critically damped since the resistance value equals the critical resistance calculated from $L$ and $C$
4. The circuit becomes underdamped with oscillation frequency $\omega_d = 2.65$ rad/s and exponential decay factor $e^{-0.75t}$ (correct answer)

Explanation: For critical damping in an RLC circuit: $R_{critical} = 2\sqrt{\frac{L}{C}} = 2\sqrt{\frac{4}{1/36}} = 2\sqrt{144} = 24$ Ω. With $R = 6$ Ω $< 24$ Ω, the circuit is underdamped. The characteristic equation is $s^2 + \frac{R}{L}s + \frac{1}{LC} = 0$, giving $s^2 + 1.5s + 9 = 0$. The discriminant is $1.5^2 - 4(9) = 2.25 - 36 = -33.75 < 0$ (underdamped). The roots are $s = -0.75 \pm i\sqrt{33.75/4} = -0.75 \pm i\sqrt{8.4375} \approx -0.75 \pm i2.65$. So $\alpha = 0.75$ and $\omega_d \approx 2.65$ rad/s.

Question 15

1. The system transitions from critically damped to underdamped because increasing $p$ makes the discriminant negative
2. The system transitions from critically damped to overdamped because the increased damping dominates the restoring force (correct answer)
3. The system remains critically damped because the 10% increase is insufficient to change the damping classification
4. The system becomes unstable because the increased $p$ value violates the stability criterion for the characteristic equation

Explanation: Initially, the system is critically damped with $\Delta = p^2 - 4q = 0$. When $p$ increases to $1.1p$, the new discriminant becomes $\Delta_{new} = (1.1p)^2 - 4q = 1.21p^2 - 4q$. Since $p^2 = 4q$ originally, we have $\Delta_{new} = 1.21(4q) - 4q = 0.84q > 0$. A positive discriminant indicates overdamped behavior with two distinct real roots. The system transitions from critical damping to overdamping because the increased damping coefficient creates two real, negative characteristic roots instead of one repeated real root.

Question 16

1. Pendulum A is critically damped and will return to equilibrium faster than pendulum B, which is underdamped and oscillates (correct answer)
2. Pendulum A is overdamped and returns to equilibrium slower than pendulum B, which is critically damped
3. Pendulum A is critically damped and pendulum B is underdamped, but pendulum B reaches equilibrium faster due to oscillatory decay
4. Both pendulums are underdamped, but pendulum A has a higher frequency of oscillation than pendulum B

Explanation: For a damped harmonic oscillator, critical damping occurs when $c = 2\sqrt{mk}$, overdamping when $c > 2\sqrt{mk}$, and underdamping when $c < 2\sqrt{mk}$. Pendulum A has $c_A = 2\sqrt{mk}$ (critically damped) and pendulum B has $c_B = \sqrt{3mk} \approx 1.73\sqrt{mk} < 2\sqrt{mk}$ (underdamped). Critical damping provides the fastest return to equilibrium without oscillation, while underdamped systems oscillate with decreasing amplitude.