Differential Equations Quiz: Mass Spring Systems

What this quiz covers

This quiz focuses on Mass Spring Systems, 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. $\sqrt{3}/\sqrt{8}$
2. 12
3. $1/(2\sqrt{3})$
4. $2\sqrt{3}$ (correct answer)

Explanation: Resonance occurs when the driving frequency $\omega$ equals the natural frequency $\omega_0 = \sqrt{k/m}$ of the system. First, find the spring constant $k$ from Hooke's Law: $W = kL \implies 16 = k(8/3)$, so $k = (16 \cdot 3)/8 = 6$ lb/ft. Next, find the mass $m$: $m = W/g = 16/32 = 1/2$ slug. Now, calculate the natural frequency: $\omega_0 = \sqrt{k/m} = \sqrt{6 / (1/2)} = \sqrt{12} = 2\sqrt{3}$ rad/s. Resonance occurs at this frequency.

Question 2

1. 2
2. 4 (correct answer)
3. 8
4. $16\sqrt{2}$

Explanation: First, determine the spring constant $k$. By Hooke's Law, the force (weight) equals $k$ times the elongation $L$. The elongation is 6 inches = 0.5 ft. So, $8 = k(0.5)$, which gives $k = 16$ lb/ft. Next, determine the mass $m$. The weight $W$ is $mg$, so $m = W/g = 8/32 = 1/4$ slug. A system is critically damped when the damping coefficient $\gamma$ satisfies the condition $\gamma^2 - 4mk = 0$. Therefore, $\gamma = \sqrt{4mk}$. Substituting the values for $m$ and $k$: $\gamma = \sqrt{4 \cdot (1/4) \cdot 16} = \sqrt{16} = 4$ lb-s/ft.

Question 3

1. The system is underdamped and unforced. (correct answer)
2. The system is overdamped and unforced.
3. The system is undamped and forced at resonance.
4. The system is underdamped and driven by an external force.

Explanation: The solution $x(t)$ has two main components. The term $e^{-0.5t}$ is an exponential decay factor, which indicates the presence of damping. The term $(3 \cos(2t) - \sin(2t))$ is a sinusoidal function, which indicates that the system is oscillating. A system that is both damped and oscillating is classified as underdamped. Since the solution decays to zero as $t \to \infty$, there is no persistent external force driving the system; if there were, a steady-state term would be present in the solution. Thus, the system is unforced (or represents the transient response of a forced system).

Question 4

1. $\sqrt{2} T$
2. $2\sqrt{2} T$ (correct answer)
3. $2T$
4. $8T$

Explanation: The period $T$ of an undamped mass-spring system is given by the formula $T = 2\pi\sqrt{m/k}$. Let the initial mass and spring constant be $m_1$ and $k_1$, so $T = 2\pi\sqrt{m_1/k_1}$. The new mass is $m_2 = 4m_1$ and the new spring constant is $k_2 = k_1/2$. The new period, $T_{new}$, is given by $T_{new} = 2\pi\sqrt{m_2/k_2} = 2\pi\sqrt{(4m_1)/(k_1/2)} = 2\pi\sqrt{8m_1/k_1} = \sqrt{8} \cdot (2\pi\sqrt{m_1/k_1}) = \sqrt{8}T = 2\sqrt{2}T$.

Question 5

1. $\omega_2 = 2\omega_1$
2. $\omega_2 = \omega_1$
3. $\omega_2 = \sqrt{2}\omega_1$ (correct answer)
4. $\omega_2 = \omega_1/2$

Explanation: The natural angular frequency is given by $\omega = \sqrt{k/m}$. For any vertical mass-spring system at equilibrium, $mg=kL$, so the spring constant is $k=mg/L$. Thus, we can express the frequency as $\omega = \sqrt{(mg/L)/m} = \sqrt{g/L}$. For System 1: $\omega_1 = \sqrt{g/L_1}$. For System 2, the elongation is $L_2 = L_1/2$. Therefore, its frequency is $\omega_2 = \sqrt{g/L_2} = \sqrt{g/(L_1/2)} = \sqrt{2g/L_1} = \sqrt{2}\sqrt{g/L_1} = \sqrt{2}\omega_1$.

Question 6

1. $t^* = \frac{\ln(10)}{3}$
2. $t^* = \frac{2\ln(10)}{3}$ (correct answer)
3. $t^* = \frac{\ln(10)}{2}$
4. $t^* = 2/3$

Explanation: The transient solution $x_c(t)$ corresponds to the homogeneous equation $x'' + 6x' + 25x = 0$. The characteristic equation is $r^2 + 6r + 25 = 0$. The roots are $r = \frac{-6 \pm \sqrt{36 - 100}}{2} = -3 \pm 4i$. The transient solution has the form $x_c(t) = e^{-3t}(c_1\cos(4t) + c_2\sin(4t))$. The amplitude of this transient solution is $A(t) = A_0 e^{-3t}$, where $A_0$ is the initial amplitude. We want to find the time $t^*$ such that $A(t^*) = 0.01 A_0$. This gives the equation $A_0 e^{-3t^*} = 0.01 A_0$, which simplifies to $e^{-3t^*} = 0.01$. Taking the natural logarithm of both sides, we get $-3t^* = \ln(0.01) = \ln(10^{-2}) = -2\ln(10)$. Solving for $t^*$ gives $t^* = \frac{2\ln(10)}{3}$.

Question 7

1. $5\pi/6$
2. $7\pi/6$
3. $5\pi/3$
4. $11\pi/6$ (correct answer)

Explanation: First, simplify the ODE to $x'' + 25x = 0$. The angular frequency is $\omega = \sqrt{25} = 5$ rad/s. The general solution is $x(t) = c_1\cos(5t) + c_2\sin(5t)$. Applying the initial condition $x(0) = \sqrt{3}$ gives $c_1 = \sqrt{3}$. The derivative is $x'(t) = -5c_1\sin(5t) + 5c_2\cos(5t)$. Applying $x'(0) = -5$ gives $5c_2 = -5$, so $c_2 = -1$. The solution is $x(t) = \sqrt{3}\cos(5t) - \sin(5t)$. To convert to phase-amplitude form $C\cos(5t - \delta) = C\cos(5t)\cos(\delta) + C\sin(5t)\sin(\delta)$, we match coefficients: $C\cos(\delta) = c_1 = \sqrt{3}$ and $C\sin(\delta) = c_2 = -1$. Thus, $\tan(\delta) = c_2/c_1 = -1/\sqrt{3}$. Since $\cos(\delta)$ is positive and $\sin(\delta)$ is negative, the angle $\delta$ must be in the fourth quadrant. The reference angle for $\tan(\theta) = 1/\sqrt{3}$ is $\pi/6$. The corresponding angle in the fourth quadrant is $\delta = 2\pi - \pi/6 = 11\pi/6$.

Question 8

1. $\sqrt{13}$
2. $2\pi/3$
3. 2
4. 3 (correct answer)

Explanation: First, write the characteristic equation for the ODE: $2r^2 + 8r + 26 = 0$. Dividing by 2 simplifies this to $r^2 + 4r + 13 = 0$. We solve for the roots using the quadratic formula: $r = \frac{-4 \pm \sqrt{4^2 - 4(1)(13)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i$. The solution to the ODE is of the form $x(t) = e^{-2t}(c_1 \cos(3t) + c_2 \sin(3t))$. The quasi-frequency is the angular frequency of the sinusoidal part, which is the imaginary part of the complex root. Therefore, the quasi-frequency is 3 rad/s.

Question 9

1. It will never pass through the equilibrium position.
2. It will pass through the equilibrium position exactly once.
3. It can pass through the equilibrium position at most once. (correct answer)
4. It will oscillate, passing through the equilibrium position infinitely many times.

Explanation: The general solution for an overdamped system is $x(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$, where $r_1$ and $r_2$ are distinct negative real roots. To find when the mass passes through equilibrium, we set $x(t)=0$. This gives $c_1 e^{r_1 t} = -c_2 e^{r_2 t}$, which can be rewritten as $e^{(r_1-r_2)t} = -c_2/c_1$. Since $r_1 \neq r_2$, the function on the left is strictly monotonic. Therefore, this equation can have at most one positive solution for $t$. If the initial conditions result in $-c_2/c_1 \le 1$, there is no positive solution. If the initial velocity is large enough and directed towards the equilibrium, it can overshoot once, but it cannot oscillate back. Thus, it can pass through equilibrium at most one time.

Question 10

1. $8/5$ (correct answer)
2. $40/7$
3. $32/25$
4. 4

Explanation: We seek a particular solution (the steady-state solution) of the form $x_p(t) = A\cos(2t) + B\sin(2t)$. We compute its derivatives: $x_p' = -2A\sin(2t) + 2B\cos(2t)$ and $x_p'' = -4A\cos(2t) - 4B\sin(2t)$. Substituting these into the ODE yields: $2(-4A\cos(2t) - 4B\sin(2t)) + 2(-2A\sin(2t) + 2B\cos(2t)) + 5(A\cos(2t) + B\sin(2t)) = 8\cos(2t)$. Grouping coefficients of $\cos(2t)$ and $\sin(2t)$ gives two equations: $(-8A + 4B + 5A) = 8 \implies -3A + 4B = 8$ and $(-8B - 4A + 5B) = 0 \implies -4A - 3B = 0$. From the second equation, $B = -4A/3$. Substituting into the first equation: $-3A + 4(-4A/3) = 8 \implies -9A/3 - 16A/3 = 8 \implies -25A/3 = 8 \implies A = -24/25$. Then $B = -4/3(-24/25) = 32/25$. The amplitude is $R = \sqrt{A^2 + B^2} = \sqrt{(-24/25)^2 + (32/25)^2} = \frac{1}{25}\sqrt{576 + 1024} = \frac{1}{25}\sqrt{1600} = \frac{40}{25} = 8/5$.

Question 11

1. 0.3 m/s
2. 1.2 m/s (correct answer)
3. $0.6/\pi$ m/s
4. 2.0 m/s

Explanation: The maximum speed of an undamped mass-spring system is given by $v_{max} = A\omega_0$, where $A$ is the amplitude and $\omega_0$ is the natural angular frequency. Since the mass is released from rest at a displacement of 0.3 m from equilibrium, the amplitude is $A = 0.3$ m. The period is given as $T = \pi/2$ s. The angular frequency is related to the period by $T = 2\pi/\omega_0$, so $\omega_0 = 2\pi/T = 2\pi/(\pi/2) = 4$ rad/s. The maximum speed is then $v_{max} = A\omega_0 = (0.3)(4) = 1.2$ m/s. The information about the mass and the initial stretch is extra information not needed to solve the problem given the period.

Question 12

1. $\sqrt{2}$, because parallel springs add their spring constants while frequency depends on $\sqrt{k/m}$ (correct answer)
2. $2$, because the effective spring constant doubles when springs are connected in parallel
3. $\frac{1}{\sqrt{2}}$, because removing a spring reduces the restoring force by half
4. $\frac{1}{2}$, because the frequency is inversely proportional to the spring constant

Explanation: For springs in parallel, the effective spring constant is $k_{eff} = k_1 + k_2 = k + k = 2k$. The natural frequency is $\omega = \sqrt{\frac{k_{eff}}{m}}$. So $\omega_1 = \sqrt{\frac{2k}{m}}$ and $\omega_2 = \sqrt{\frac{k}{m}}$. Therefore, $\frac{\omega_1}{\omega_2} = \frac{\sqrt{2k/m}}{\sqrt{k/m}} = \sqrt{2}$. Choice B confuses the spring constant ratio with the frequency ratio. Choice C inverts the relationship. Choice D incorrectly states that frequency is inversely proportional to spring constant.

Question 13

1. Remains the same, since the spring constant is unchanged and determines the total energy
2. Increases by a factor of 8, since both mass and amplitude changes affect the kinetic energy component
3. Increases by a factor of 2, since the mass change compensates for half of the amplitude effect
4. Increases by a factor of 4, since energy is proportional to amplitude squared regardless of mass change (correct answer)

Explanation: When you encounter spring-mass oscillation problems, remember that two key relationships govern the system: the period depends on both mass and spring constant, while mechanical energy depends on amplitude and spring properties. First, let's find the spring constant from the original setup. For a spring-mass system, $T = 2\pi\sqrt{\frac{m}{k}}$. With $m_1 = 2$ kg and $T = \frac{2\pi}{3}$: $\frac{2\pi}{3} = 2\pi\sqrt{\frac{2}{k}}$ Solving: $\frac{1}{3} = \sqrt{\frac{2}{k}}$, so $k = 18$ N/m. The total mechanical energy of a spring-mass system is $E = \frac{1}{2}kA^2$, where $A$ is the amplitude. Notice that energy depends only on the spring constant and amplitude — mass doesn't appear in this formula. In the new scenario, the spring constant remains 18 N/m, but the amplitude doubles. Therefore: $E_{new} = \frac{1}{2}k(2A)^2 = \frac{1}{2}k(4A^2) = 4 \cdot \frac{1}{2}kA^2 = 4E_{original}$ The energy increases by a factor of 4. Choice A incorrectly assumes energy depends only on the spring constant. Choice B mistakenly includes mass in the energy calculation and arrives at an incorrect factor. Choice C suggests some compensation effect between mass and amplitude changes, but since mass doesn't affect total energy, this reasoning is flawed. Choice D correctly recognizes that energy is proportional to amplitude squared, regardless of mass changes. Study tip: Remember that for spring systems, total mechanical energy depends only on $k$ and $A^2$ — the mass affects period and frequency, but not total energy.

Question 14

1. $-1.6$ m/s², calculated from the spring force at the given displacement position
2. $-2.4$ m/s², directed toward equilibrium according to Hooke's law restoring force (correct answer)
3. $+2.4$ m/s², directed away from equilibrium due to the oscillatory motion phase
4. $-4.0$ m/s², found by considering the maximum restoring force capability

Explanation: When you encounter a spring-mass system problem, you're dealing with simple harmonic motion governed by Hooke's law. The key insight is that the acceleration depends only on the current displacement, not the motion history. At any position in simple harmonic motion, the restoring force follows Hooke's law: $F = -kx$, where the negative sign indicates the force opposes displacement. Using Newton's second law, $F = ma$, we get $ma = -kx$, so $a = -\frac{kx}{m}$. At the given displacement of 0.15 m: $a = -\frac{16 \times 0.15}{1} = -2.4$ m/s². The negative sign confirms the acceleration points toward equilibrium (opposite to the displacement direction), which is exactly what Hooke's law predicts for a restoring force. Looking at the wrong answers: Choice A uses the correct approach but makes a calculation error ($16 \times 0.15 = 2.4$, not 1.6). Choice C has the right magnitude but wrong sign—it incorrectly suggests acceleration points away from equilibrium, which would make the system unstable rather than oscillatory. Choice D calculates acceleration using the maximum displacement (0.25 m) instead of the current position (0.15 m), showing a fundamental misunderstanding that acceleration changes with position. Study tip: In spring problems, always remember that acceleration equals $-\frac{k}{m}$ times the current displacement. The initial conditions only matter for finding velocity or determining the oscillation amplitude—acceleration depends solely on where the mass is right now.

Question 15

1. $t = \frac{5\pi}{12}$ seconds, considering both the frequency and phase shift of oscillation
2. $t = \frac{\pi}{6}$ seconds, accounting for the phase shift in the cosine function
3. $t = \frac{\pi}{12}$ seconds, when the cosine term equals zero for the first time (correct answer)
4. $t = \frac{\pi}{4}$ seconds, when the phase angle makes the displacement zero

Explanation: When analyzing damped oscillations, you need to find when the entire displacement function equals zero, not just individual components. The position function $x(t) = 0.2e^{-0.5t}\cos(3t + \frac{\pi}{4})$ represents a damped harmonic oscillator where the exponential term controls amplitude decay and the cosine term controls oscillation. To find when the mass crosses equilibrium ($x = 0$), set the entire function equal to zero: $0.2e^{-0.5t}\cos(3t + \frac{\pi}{4}) = 0$. Since the exponential term $e^{-0.5t}$ is always positive and never zero, and the constant 0.2 is non-zero, the only way this product can equal zero is if $\cos(3t + \frac{\pi}{4}) = 0$. Cosine equals zero when its argument equals $\frac{\pi}{2} + n\pi$ for integer $n$. For the first occurrence after $t = 0$, set $3t + \frac{\pi}{4} = \frac{\pi}{2}$. Solving: $3t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$, so $t = \frac{\pi}{12}$ seconds. Choice A ($\frac{5\pi}{12}$) incorrectly adds rather than subtracts the phase shift. Choice B ($\frac{\pi}{6}$) uses the wrong approach, possibly confusing the angular frequency with the phase calculation. Choice D ($\frac{\pi}{4}$) forgets to divide by the angular frequency coefficient of 3. Remember: for damped oscillations, equilibrium crossings occur when the oscillatory term (sine or cosine) equals zero, regardless of the damping factor. Always solve the trigonometric equation completely, accounting for both phase shifts and frequency scaling.

Question 16

1. $\frac{10}{9}$ m, found by solving the particular solution for the non-homogeneous equation (correct answer)
2. $\frac{2}{5}$ m, found by applying the resonance frequency formula with damping correction
3. $\frac{10}{41}$ m, found by considering both the spring restoring force and driving frequency
4. $\frac{5}{8}$ m, found by using the amplitude formula for undamped forced oscillations

Explanation: For the undamped forced oscillator $mx'' + kx = F_0\cos(\omega t)$, the steady-state solution has amplitude $A = \frac{F_0}{|k - m\omega^2|}$. Here, $F_0 = 10$, $\omega = 4$, $m = 1$, $k = 25$. So $A = \frac{10}{|25 - 1(16)|} = \frac{10}{|25 - 16|} = \frac{10}{9}$ m. Choice B incorrectly applies resonance formulas. Choice C uses an incorrect denominator calculation. Choice D doesn't properly account for the relationship between driving frequency and natural frequency.