Differential Equations Quiz: Forced Oscillations And Resonance

What this quiz covers

This quiz focuses on Forced Oscillations And Resonance, 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. $4$ rad/s
2. $2\sqrt{3}$ rad/s
3. $2\sqrt{2}$ rad/s (correct answer)
4. $2$ rad/s

Explanation: The amplitude of the steady-state solution is maximized at the resonant frequency $\omega_r$. For a system $my'' + \gamma y' + ky = F_0 \cos(\omega t)$, the natural frequency is $\omega_0 = \sqrt{k/m}$ and the resonant frequency is $\omega_r = \sqrt{\omega_0^2 - \frac{\gamma^2}{2m^2}}$. Given $m=2$, $\gamma=8$, and $k=32$, the natural frequency squared is $\omega_0^2 = k/m = 32/2 = 16$. So, $\omega_0 = 4$ rad/s. The resonant frequency is $\omega_r = \sqrt{16 - \frac{8^2}{2 \cdot 2^2}} = \sqrt{16 - \frac{64}{8}} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$ rad/s. (A) This is the natural frequency $\omega_0 = \sqrt{k/m} = 4$. Resonance in a damped system occurs at a frequency slightly lower than the natural frequency. (B) This is the damped natural frequency $\omega_d = \sqrt{\omega_0^2 - (\gamma/2m)^2} = \sqrt{16 - (8/4)^2} = \sqrt{12} = 2\sqrt{3}$. This determines the frequency of the transient solution, not the peak of the forced response. (D) This value could result from various calculation errors, such as misplacing a factor of 2 in the resonant frequency formula.

Question 2

1. $\gamma=0$ and $\omega=3$ (correct answer)
2. $\gamma > 0$ and $\omega=3$
3. $\gamma=0$ and any $\omega > 0$
4. $\gamma>0$ and $\omega$ is sufficiently close to $3$

Explanation: Unbounded solutions due to resonance occur only when two conditions are met: (1) there is no damping in the system, and (2) the driving frequency matches the natural frequency of the system. The given equation has $m=1$ and $k=9$, so the natural angular frequency is $\omega_0 = \sqrt{k/m} = \sqrt{9/1} = 3$ rad/s. For resonance to produce an unbounded solution, the damping coefficient must be zero ($\gamma=0$) and the driving frequency must equal the natural frequency ($\omega = \omega_0 = 3$). (B) If there is any damping ($\gamma > 0$), the amplitude of the steady-state solution is always finite, even when $\omega = \omega_0$. The damping term dissipates energy, preventing unbounded growth. (C) If the system is undamped ($\gamma=0$), forcing it at a frequency $\omega$ that does not match the natural frequency $\omega_0$ will result in a bounded solution exhibiting beats, not unbounded growth. (D) This describes the condition for practical resonance in a damped system, which results in a large but still bounded amplitude, not an unbounded solution.

Question 3

1. The solution will oscillate with an amplitude that grows indefinitely over time.
2. The solution will approach a sinusoidal oscillation with a constant amplitude and an angular frequency of $2$ rad/s. (correct answer)
3. The solution will decay to zero because of the positive damping term.
4. The long-term behavior of the solution will be a large-amplitude oscillation determined by the initial conditions.

Explanation: The general solution is $y(t) = y_c(t) + y_p(t)$. The homogeneous equation $y'' + 0.1y' + 4y = 0$ has a characteristic equation $r^2 + 0.1r + 4 = 0$. The roots have a negative real part, so the transient solution $y_c(t)$ decays to zero as $t \to \infty$. The long-term behavior is therefore governed by the steady-state (particular) solution $y_p(t)$. The forcing function is $2\cos(2t)$, so the system is driven at an angular frequency of $\omega=2$ rad/s. The natural frequency is $\omega_0 = \sqrt{4}=2$ rad/s. Since the system is damped ($\gamma=0.1 > 0$) and driven at its natural frequency, it will exhibit a large but finite amplitude. The steady-state solution will be a sinusoidal oscillation at the driving frequency, $\omega=2$ rad/s. (A) Unbounded amplitude only occurs in an undamped system ($\gamma=0$) driven at its natural frequency. (C) The forcing term prevents the solution from decaying to zero; it sustains a steady-state oscillation. (D) The initial conditions determine the constants in the transient solution $y_c(t)$, which decays to zero and does not affect the long-term (steady-state) behavior.

Question 4

1. The system is in resonance, with $\omega_0 = 10.5$ rad/s and $F(t)$ having a frequency of $10.5$ rad/s.
2. The natural frequency is $\omega_0 = 10$ rad/s and the forcing frequency is $11$ rad/s, or vice versa. (correct answer)
3. The natural frequency is $\omega_0 = 0.5$ rad/s and the forcing frequency is $10.5$ rad/s.
4. The system is overdamped, and the forcing function is a sum of two sinusoids.

Explanation: The given solution form represents the phenomenon of beats, which occurs when a system is driven at a frequency close to its natural frequency. Using the product-to-sum trigonometric identity $2\sin(A)\sin(B) = \cos(A-B) - \cos(A+B)$, we can rewrite the solution as: $y(t) = 4 \cdot [2\sin(10.5t)\sin(0.5t)] = 4[\cos(10.5t - 0.5t) - \cos(10.5t + 0.5t)] = 4[\cos(10t) - \cos(11t)]$. This solution is a superposition of two oscillations with angular frequencies $10$ rad/s and $11$ rad/s. In a forced, undamped system initially at rest, the solution has the form $C(\cos(\omega t) - \cos(\omega_0 t))$, where $\omega$ is the driving frequency and $\omega_0$ is the natural frequency. Therefore, the two frequencies involved must be $\omega_0=10$ and $\omega=11$, or $\omega_0=11$ and $\omega=10$. (A) Resonance in an undamped system leads to a solution with an amplitude that grows linearly with $t$, of the form $At\sin(\omega_0 t)$, not beats. (C) This confuses the beat frequency and the average frequency with the actual system frequencies. (D) The problem states the system is undamped, so it cannot be overdamped. The form of the solution is characteristic of a single sinusoidal forcing function.

Question 5

1. $0$
2. $\pi/4$
3. $\pi/2$ (correct answer)
4. $\pi$

Explanation: For a system $my'' + \gamma y' + ky = F_0 \cos(\omega t)$, the phase shift $\delta$ of the steady-state solution is given by the formula $\tan(\delta) = \frac{\gamma \omega}{k - m\omega^2}$. In this problem, $m=1$, $\gamma>0$, and $k=25$. The natural frequency is $\omega_0 = \sqrt{k/m} = \sqrt{25}=5$ rad/s. The formula becomes $\tan(\delta) = \frac{\gamma \omega}{25 - \omega^2}$. As the driving frequency $\omega$ approaches the natural frequency $\omega_0 = 5$, the denominator $25 - \omega^2$ approaches $0$. The numerator $\gamma \omega$ approaches $5\gamma$, which is a positive constant. Therefore, $\tan(\delta)$ approaches $+\infty$ (if approaching from $\omega < 5$). The angle whose tangent approaches $+\infty$ is $\pi/2$. This means the displacement response is 90 degrees out of phase with the driving force at resonance. (A) The phase shift approaches $0$ as the driving frequency $\omega$ approaches $0$. (B) There is no special significance to $\pi/4$ in this context. (D) The phase shift approaches $\pi$ as the driving frequency $\omega$ approaches $\infty$.

Question 6

1. $4\sqrt{5}$
2. $8$
3. $2\sqrt{5}$
4. $2\sqrt{10}$ (correct answer)

Explanation: This problem tests your understanding of resonance in damped harmonic oscillators. When you see a differential equation of the form $my'' + \gamma y' + ky = F(t)$ with a driving force, you're dealing with forced oscillations where amplitude depends on the driving frequency. For maximum steady-state amplitude in a damped system, resonance occurs at the frequency $\omega_{\text{res}} = \sqrt{\omega_0^2 - \frac{\gamma^2}{2m^2}}$, where $\omega_0 = \sqrt{\frac{k}{m}}$ is the natural frequency. From your equation $2y'' + \gamma y' + 18y = F(t)$, we have $m = 2$ and $k = 18$, so $\omega_0 = \sqrt{\frac{18}{2}} = 3$ rad/s. Since resonance occurs at $\omega = 2$ rad/s, we can solve: $2 = \sqrt{9 - \frac{\gamma^2}{8}}$ Squaring both sides: $4 = 9 - \frac{\gamma^2}{8}$ Rearranging: $\frac{\gamma^2}{8} = 5$ Therefore: $\gamma^2 = 40$, so $\gamma = 2\sqrt{10}$ Answer choice A ($4\sqrt{5}$) would give $\gamma^2 = 80$, making the resonant frequency too low. Choice B (8) gives $\gamma^2 = 64$, also incorrect. Choice C ($2\sqrt{5}$) gives $\gamma^2 = 20$, which would place resonance at a higher frequency than 2 rad/s. The correct answer is D. Study tip: Always identify the natural frequency first, then use the resonance condition. Remember that damping shifts the resonant frequency below the natural frequency, and heavier damping means a larger shift.

Question 7

1. $A_1(\omega) > A_2(\omega)$ for all $\omega > 0$.
2. The relationship between $A_1(\omega)$ and $A_2(\omega)$ cannot be determined without knowing the value of $F_0$.
3. $A_1(\omega) < A_2(\omega)$ only when $\omega$ is near the natural frequency, and the inequality reverses otherwise.
4. $A_1(\omega) < A_2(\omega)$ for all $\omega > 0$. (correct answer)

Explanation: When you encounter forced oscillation problems comparing different damping coefficients, focus on how damping affects the steady-state amplitude response across all frequencies. For a driven harmonic oscillator $y'' + 2\gamma y' + \omega_0^2 y = F_0 \cos(\omega t)$, the steady-state amplitude is $A(\omega) = \frac{F_0}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}}$ Both systems have the same natural frequency $\omega_0 = 2$ (since both have coefficient 4 for the $y$ term), but different damping: System 1 has $\gamma_1 = 0.25$ while System 2 has $\gamma_2 = 0.1$. Since $\gamma_1 > \gamma_2$, System 1 is more heavily damped. The key insight is that higher damping always reduces amplitude at every frequency. The denominator $\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}$ increases with larger $\gamma$, making the amplitude smaller. This holds universally across the frequency spectrum. Choice A is backwards - the more damped system (System 1) has smaller amplitude. Choice B incorrectly suggests $F_0$ matters for the comparison; since $F_0$ appears in both numerators, it cancels out when comparing ratios. Choice C suggests the relationship reverses at different frequencies, but this never happens - less damping always means higher amplitude regardless of whether you're near resonance or far from it. Study tip: Remember that damping always suppresses motion. In amplitude comparisons for identical systems with different damping, the less damped system always wins at every frequency - there's no crossover point.

Question 8

1. $\frac{\sqrt{10}}{2}$ A
2. $\sqrt{10}$ A (correct answer)
3. $5\sqrt{2}$ A
4. $\frac{10}{3}$ A

Explanation: The amplitude of the steady-state current can be found using the concept of impedance in an AC circuit. The impedance $Z$ is given by $Z = R + i(\omega L - \frac{1}{\omega C})$. The amplitude of the current, $|I|$, is then $|E_0| / |Z|$. First, calculate the impedance: $Z = 3 + i(2 \cdot 1 - \frac{1}{2 \cdot (1/2)}) = 3 + i(2 - 1) = 3 + i$. Next, find the magnitude of the impedance: $|Z| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$ $\Omega$. Finally, the amplitude of the steady-state current is: $|I| = \frac{E_0}{|Z|} = \frac{10}{\sqrt{10}} = \sqrt{10}$ A. (A) This is the amplitude of the steady-state charge $Q(t)$, not the current $I(t)$. The amplitude of the current is $\omega$ times the amplitude of the charge. (C) This value does not correspond to a standard calculation for this circuit and likely arises from a conceptual error. (D) This would be the amplitude if the impedance were purely resistive ($|Z|=R=3$), ignoring the effects of the inductor and capacitor.

Question 9

1. $1$ rad/s (correct answer)
2. $2$ rad/s
3. $101$ rad/s
4. $202$ rad/s

Explanation: The phenomenon of beats arises from the superposition of two oscillations with close frequencies, in this case the natural frequency $\omega_0=100$ and the driving frequency $\omega=102$. The resulting motion can be expressed as a product of two sinusoids: a rapid oscillation and a slow oscillation (the envelope). The angular frequency of the rapid oscillation is the average of the two frequencies, $(\omega_0+\omega)/2$. The angular frequency of the envelope (the beat frequency) is half the difference of the two frequencies, $|\omega_0-\omega|/2$. Beat angular frequency = $\frac{|100 - 102|}{2} = \frac{2}{2} = 1$ rad/s. (B) This is the absolute difference between the frequencies, $|\omega_0-\omega|$. The envelope frequency is half of this value. (C) This is the frequency of the rapid oscillation, $(\omega_0+\omega)/2 = (100+102)/2 = 101$ rad/s. (D) This is the sum of the two frequencies, which is not physically represented as a distinct frequency in the beat phenomenon.

Question 10

1. The amplitude will increase. (correct answer)
2. The amplitude will decrease.
3. The amplitude will first increase, then decrease.
4. The effect depends on whether $\omega$ is greater or less than the natural frequency.

Explanation: The amplitude of the steady-state solution is given by $A(\gamma) = \frac{F_0}{\sqrt{m^2(\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2}}$. In this problem, $m=1, k=100$, so $\omega_0^2 = 100$. The amplitude is $A(\gamma) = \frac{1}{\sqrt{(100 - \omega^2)^2 + \gamma^2 \omega^2}}$. The driving frequency $\omega$ is fixed. As the damping coefficient $\gamma$ decreases, the term $\gamma^2 \omega^2$ in the denominator gets smaller. Since all other terms in the denominator are constant, the value of the entire denominator decreases. Consequently, the amplitude $A(\gamma)$, which is the reciprocal of the square root of the denominator, must increase. (B) The amplitude would decrease if $\gamma$ were increasing. (C) The behavior of first increasing and then decreasing describes the amplitude as a function of the driving frequency $\omega$ (the resonance curve), not as a function of the damping coefficient $\gamma$. (D) The term involving $\omega$ and $\omega_0$ is squared, $(100 - \omega^2)^2$, so its value is always non-negative regardless of whether $\omega > \omega_0$ or $\omega < \omega_0$. Thus, decreasing $\gamma$ will always increase the amplitude for any fixed $\omega ≠ 0$.

Question 11

1. $\omega_r(\gamma)$ increases and $A_{max}(\gamma)$ decreases.
2. $\omega_r(\gamma)$ decreases and $A_{max}(\gamma)$ increases.
3. $\omega_r(\gamma)$ decreases and $A_{max}(\gamma)$ decreases. (correct answer)
4. $\omega_r(\gamma)$ remains constant and $A_{max}(\gamma)$ decreases.

Explanation: The resonant frequency is given by $\omega_r(\gamma) = \sqrt{\omega_0^2 - \frac{\gamma^2}{2m^2}}$, where $\omega_0 = \sqrt{k/m}$. As the damping coefficient $\gamma$ increases, the term $\frac{\gamma^2}{2m^2}$ increases, causing the entire expression under the square root to decrease. Thus, $\omega_r(\gamma)$ decreases. The amplitude of the steady-state solution is $A(\omega) = \frac{F_0}{\sqrt{m^2(\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2}}$. Greater damping (larger $\gamma$) increases the denominator for any given $\omega$, which reduces the amplitude across all frequencies. Therefore, the maximum amplitude, $A_{max}(\gamma)$, also decreases. (A) $\omega_r(\gamma)$ decreases, it does not increase. (B) $A_{max}(\gamma)$ decreases, it does not increase. (D) The resonant frequency $\omega_r$ is dependent on $\gamma$ and is not constant, unless there is no damping. The misconception is that resonance always occurs at the natural frequency $\omega_0$.

Question 12

1. $R = \sqrt{7}$, $\alpha = \arctan(3/4)$; the solution exhibits beating between the two frequency components
2. $R = 7$, $\alpha = \arctan(4/3)$; the solution amplitude is larger due to the combined forcing terms
3. $R = 5$, $\alpha = \arctan(4/3)$; the solution requires separate particular solutions for each term
4. $R = 5$, $\alpha = \arctan(3/4)$; the solution has the same form but shifted phase reference (correct answer)

Explanation: When you encounter a forced oscillator with multiple trigonometric terms, the key insight is that any combination of sine and cosine functions with the same frequency can be rewritten as a single sinusoidal function with appropriate amplitude and phase shift. To convert $4\cos(\omega t) + 3\sin(\omega t)$ into $R\cos(\omega t - \alpha)$, you need to find the resultant amplitude and phase. The amplitude is $R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$. For the phase shift, use the identity: $a\cos(\omega t) + b\sin(\omega t) = R\cos(\omega t - \alpha)$ where $\tan(\alpha) = b/a$. Here, $\tan(\alpha) = 3/4$, so $\alpha = \arctan(3/4)$. The beauty of this transformation is that it doesn't change the physics—you still have a single-frequency driving force, just expressed more conveniently. The steady-state solution will have the same mathematical form as if you started with a simple cosine drive, but with this combined amplitude and the phase reference shifted by $\alpha$. Choice A incorrectly calculates $R = \sqrt{7}$ and mentions beating, which only occurs with multiple frequencies. Choice B gives the wrong values ($R = 7$ and $\alpha = \arctan(4/3)$) and misunderstands the amplitude relationship. Choice C has the correct $R = 5$ but wrong phase, plus incorrectly suggests you need separate particular solutions—the combined form is actually more efficient. Remember: when combining trigonometric functions of the same frequency, always check if you can simplify using amplitude-phase form before solving the differential equation.

Question 13

1. $x_{new}(t) = A\cos(3t - \pi/6 + \phi + \pi/2)$; the phase relationship depends on the system's natural frequency
2. $x_{new}(t) = A\cos(3t + \pi/3 + \phi)$; the amplitude changes but the relative phase lag remains $\phi$
3. $x_{new}(t) = A\cos(3t + \pi/3 + \phi)$; both amplitude and phase lag are modified due to coupling effects
4. $x_{new}(t) = A\cos(3t + \pi/3 + \phi)$; the solution phase shifts by the same amount as the input (correct answer)

Explanation: When you encounter forced oscillation problems with sinusoidal driving terms, the key principle is linearity: how the system responds to phase changes in the input directly translates to the output. For the differential equation $\ddot{x} + 4\dot{x} + 13x = F(t)$, the steady-state response to any sinusoidal forcing depends on the system's transfer function, which determines both amplitude scaling and phase lag. The crucial insight is that when you shift the input phase, the output phase shifts by exactly the same amount while maintaining the same amplitude and relative phase relationship. The original driving force $26\cos(3t - \pi/6)$ produces $x_{ss}(t) = A\cos(3t - \pi/6 + \phi)$. When the input becomes $26\cos(3t + \pi/3)$, you've shifted the input phase by $\pi/3 - (-\pi/6) = \pi/2$. The system responds with the same phase shift: $x_{new}(t) = A\cos(3t + \pi/3 + \phi)$. The amplitude $A$ and phase lag $\phi$ remain unchanged because they depend only on the driving frequency (3 rad/s) and system parameters, not the absolute phase. Answer A incorrectly adds an extra $\pi/2$ and wrongly suggests the natural frequency matters for steady-state response. Answer B correctly gives the form but incorrectly claims amplitude changes. Answer C also has the right form but wrongly suggests coupling effects modify both amplitude and phase lag. Study tip: For linear systems, input phase shifts translate directly to output phase shifts of the same magnitude. The system's amplitude and phase lag characteristics depend only on frequency, not absolute phase.

Question 14

1. The response exhibits beating with frequency $(\omega_2 - \omega_1)/2$ and the amplitude envelope oscillates symmetrically
2. The response shows beating, but the envelope is asymmetric due to different amplitudes at $\omega_1$ and $\omega_2$ (correct answer)
3. No beating occurs because the frequencies are in the steady-state regime; the response is simply a sum of two separate responses
4. The system locks onto the frequency closer to $\omega_{res}$ and the other component gradually disappears

Explanation: When driven by two close frequencies, the system responds to each component according to its frequency response function. Since $\omega_1 < \omega_{res} < \omega_2$, the amplitudes $A_1$ and $A_2$ are different (one is closer to resonance). The total response is $x(t) = A_1\cos(\omega_1 t + \phi_1) + A_2\cos(\omega_2 t + \phi_2)$, which can be rewritten to show beating with envelope frequency $(\omega_2 - \omega_1)/2$. However, since $A_1 \neq A_2$, the envelope doesn't go to zero—it oscillates between $|A_2 - A_1|$ and $A_1 + A_2$, creating asymmetric beating. Choice A assumes equal amplitudes. Choice C ignores the beating phenomenon. Choice D describes nonlinear behavior not present in linear systems.

Question 15

1. $\omega = 3$, the natural frequency where amplitude changes are symmetric
2. $\omega = \sqrt{8.28} \approx 2.88$, the frequency of maximum amplitude response (correct answer)
3. $\omega = 0$, where the static deflection determines the amplitude behavior
4. $\omega = \sqrt{9.72} \approx 3.12$, corresponding to the inflection point of the response curve

Explanation: To find where $dA/d\omega = 0$, we differentiate the amplitude function. The maximum occurs when the denominator is minimized, which happens when $\frac{d}{d\omega}[(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2] = 0$. This gives $2(\omega_0^2 - \omega^2)(-2\omega) + 2(2\gamma\omega)(2\gamma) = 0$, which simplifies to $\omega(\omega_0^2 - \omega^2) = 2\gamma^2\omega$. For $\omega \neq 0$, this yields $\omega^2 = \omega_0^2 - 2\gamma^2 = 9 - 0.72 = 8.28$, so $\omega = \sqrt{8.28} \approx 2.88$. Choice A gives the natural frequency, not the resonance frequency. Choice C is trivial. Choice D doesn't correspond to any critical point.

Question 16

1. $\omega = 2 \pm 0.25$, corresponding to the half-power bandwidth frequencies (correct answer)
2. $\omega = 2 \pm 0.5$, where the phase difference is $\pm \pi/4$ from resonance
3. $\omega = \sqrt{3.75}$ or $\omega = \sqrt{4.25}$, the frequencies where amplitude is $1/\sqrt{2}$ times maximum
4. $\omega = 1.5$ or $\omega = 2.5$, representing the critical damping transition points

Explanation: Power absorption is proportional to $\gamma \omega^2 A^2$ where $A$ is the amplitude. Maximum power occurs at $\omega = \omega_0 = 2$. Half-power points occur when the amplitude is $1/\sqrt{2}$ times the maximum amplitude, which happens at $\omega = \omega_0 \pm \gamma/2 = 2 \pm 0.25$. Choice B uses the wrong bandwidth (should be $\gamma$, not $2\gamma$). Choice C gives the correct amplitude condition but wrong frequency calculation. Choice D refers to irrelevant damping concepts.

Question 17

1. $\omega = \sqrt{5}$ and the amplitude is maximized at this frequency
2. $\omega = \sqrt{3}$ and the amplitude grows without bound over time
3. $\omega = \sqrt{3}$ and the amplitude is maximized but remains finite (correct answer)
4. $\omega = \sqrt{5}$ and the amplitude oscillates with a beating pattern

Explanation: For a damped driven oscillator $x'' + 2\gamma x' + \omega_0^2 x = F_0\cos(\omega t)$, we have $\omega_0^2 = 5$ and $\gamma = 1$. The resonance frequency for maximum amplitude is $\omega_{res} = \sqrt{\omega_0^2 - 2\gamma^2} = \sqrt{5 - 2} = \sqrt{3}$. At this frequency, the amplitude is maximized but finite due to damping. Choice A uses the natural frequency instead of resonance frequency. Choice B incorrectly suggests unbounded growth (only occurs in undamped systems). Choice D describes beating, which occurs with nearby frequencies, not resonance.