Forced Oscillations & Resonance - Differential Equations

A forced, damped mass-spring system is described by the differential equation $2y'' + 8y' + 32y = 4\cos(\omega t)$ . At which of the following driving angular frequencies $\omega$ is the amplitude of the steady-state solution maximized?
Answer: $2\sqrt{2}$ rad/s 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.
A forced harmonic oscillator has the equation $x'' + \gamma x' + \omega_0^2 x = F_0\cos(\omega t)$ where $\gamma = 0.5$ , $\omega_0 = 2$ , and $F_0 = 8$ . At what driving frequency $\omega$ will the power absorbed by the system be exactly half of the maximum possible power absorption?
Answer: $\omega = 2 \pm 0.25$ , corresponding to the half-power bandwidth frequencies 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.
Consider a forced oscillator described by $\ddot{y} + 3\dot{y} + 2y = 4\cos(\omega t) + 3\sin(\omega t)$ . The driving force can be written as $F(t) = R\cos(\omega t - \alpha)$ . What are the values of $R$ and $\alpha$ , and how does this affect the steady-state solution compared to a simple cosine drive?
Answer: $R = 5$ , $\alpha = \arctan(3/4)$ ; the solution has the same form but shifted phase reference 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.
A mass-spring system with damping is driven by an external force $F(t) = 4\cos(\omega t)$ . The equation of motion is $x'' + 2x' + 5x = 4\cos(\omega t)$ . If the system exhibits resonance, what is the driving frequency $\omega$ and what characterizes the steady-state amplitude behavior?
Answer: $\omega = \sqrt{3}$ and the amplitude is maximized but remains finite 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.
Consider the initial value problem $y'' + 0.1y' + 4y = 2\cos(2t)$ , with $y(0)=10$ and $y'(0)=0$ . Which of the following best describes the behavior of the solution $y(t)$ for large values of $t$ ?
Answer: The solution will approach a sinusoidal oscillation with a constant amplitude and an angular frequency of $2$ rad/s. 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.
The response amplitude $A(\omega)$ of a driven damped oscillator is given by $A(\omega) = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (2\gamma\omega)^2}}$ . If $\omega_0 = 3$ and $\gamma = 0.6$ , at which frequency is the rate of change of amplitude with respect to frequency, $dA/d\omega$ , equal to zero?
Answer: $\omega = \sqrt{8.28} \approx 2.88$ , the frequency of maximum amplitude response 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.
Consider two distinct mass-spring systems, System 1 and System 2, both driven by the same forcing function $F(t) = F_0 \cos(\omega t)$ . System 1: $y_1'' + 0.5 y_1' + 4y_1 = F_0 \cos(\omega t)$ System 2: $y_2'' + 0.2 y_2' + 4y_2 = F_0 \cos(\omega t)$ Let $A_1(\omega)$ and $A_2(\omega)$ be the amplitudes of the steady-state solutions for System 1 and System 2, respectively. Which statement is true for all driving frequencies $\omega > 0$ ?
Answer: $A_1(\omega) < A_2(\omega)$ for all $\omega > 0$ . 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.
An undamped oscillator with a natural angular frequency of $\omega_0 = 100$ rad/s is subjected to a forcing function $F(t) = F_0\cos(102t)$ . The system is initially at rest. The resulting motion exhibits beats. What is the angular frequency of the slow oscillation (the envelope)?
Answer: $1$ rad/s 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.
A mass-spring system with forcing is modeled by $my'' + \gamma y' + ky = F_0 \cos(\omega t)$ , where $m, k, F_0$ are fixed positive constants. Assume the damping coefficient $\gamma$ is small enough that a resonant peak exists. Let $\omega_r(\gamma)$ be the resonant frequency and $A_{max}(\gamma)$ be the maximum amplitude of the steady-state solution. If $\gamma$ is increased, what is the effect on $\omega_r(\gamma)$ and $A_{max}(\gamma)$ ?
Answer: $\omega_r(\gamma)$ decreases and $A_{max}(\gamma)$ decreases. 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 $\omega_0$ .
For the equation $\ddot{x} + 4\dot{x} + 13x = 26\cos(3t - \pi/6)$ , the steady-state solution is $x_{ss}(t) = A\cos(3t - \pi/6 + \phi)$ where $\phi$ is the phase lag. If the driving force phase is changed to $26\cos(3t + \pi/3)$ , how does the new steady-state solution $x_{new}(t)$ relate to the original?
Answer: $x_{new}(t) = A\cos(3t + \pi/3 + \phi)$ ; the solution phase shifts by the same amount as the input 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.
A system is described by $y'' + \gamma y' + 25y = \cos(\omega t)$ , where $\gamma>0$ is a small damping coefficient. The steady-state solution can be written as $y_p(t) = A(\omega) \cos(\omega t - \delta(\omega))$ . What value does the phase shift $\delta(\omega)$ approach as the driving frequency $\omega$ approaches the natural frequency $\omega_0 = 5$ ?
Answer: $\pi/2$ 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 $\infty$ .
A damped mechanical system is modeled by $2y'' + \gamma y' + 18y = F(t)$ . It is observed that the amplitude of the steady-state response is maximized when the driving angular frequency is $\omega = 2$ rad/s. What is the value of the damping coefficient $\gamma$ ?
Answer: $2\sqrt{10}$ 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.
An RLC circuit is modeled by $L Q'' + R Q' + \frac{1}{C} Q = E_0 \sin(\omega t)$ . Given $L=1$ H, $R=3$ $\Omega$ , and $C=1/2$ F, what is the amplitude of the steady-state current $I(t) = Q'(t)$ if the input voltage has amplitude $E_0=10$ V and angular frequency $\omega=2$ rad/s?
Answer: $\sqrt{10}$ A 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.
Consider the differential equation $y'' + \gamma y' + 9y = \sin(\omega t)$ . For which combination of non-negative parameter values $\gamma$ and $\omega$ will solutions be unbounded as $t \to \infty$ ?
Answer: $\gamma=0$ and $\omega=3$ 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.
A mass of $m=1$ kg is attached to a spring with constant $k=100$ N/m. The system is damped with coefficient $\gamma$ and driven by a force $F(t) = \cos(\omega t)$ . The driving frequency $\omega$ is fixed at a value that is not the natural frequency. If the damping coefficient $\gamma$ is decreased from a large value towards zero (but remains positive), what is the effect on the amplitude of the steady-state oscillation?
Answer: The amplitude will increase. 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 $\omega ≠ 0$ .
A mass attached to a spring and dashpot is driven by $F(t) = F_0[\cos(\omega_1 t) + \cos(\omega_2 t)]$ where $\omega_1$ and $\omega_2$ are close but not equal. The system parameters give resonance frequency $\omega_{res}$ such that $\omega_1 < \omega_{res} < \omega_2$ . Which statement best describes the long-term behavior?
Answer: The response shows beating, but the envelope is asymmetric due to different amplitudes at $\omega_1$ and $\omega_2$ 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.
The solution to an initial value problem for a forced, undamped mass-spring system, $y'' + \omega_0^2 y = F(t)$ , is found to be $y(t) = 8\sin(0.5t)\sin(10.5t)$ . Which of the following can be inferred about the forcing function $F(t)$ and the natural angular frequency $\omega_0$ ?
Answer: The natural frequency is $\omega_0 = 10$ rad/s and the forcing frequency is $11$ rad/s, or vice versa. 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.

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