Amplitude, Frequency & Phase - Differential Equations

1. A critically damped oscillator has the solution $x(t) = (A + Bt)e^{-\gamma t}$. If this system is modified to become underdamped with the same natural frequency $\omega_0$ but damping coefficient reduced by half, and the resulting oscillatory motion has the same initial amplitude $A$, what is the ratio of the new oscillation frequency to the original critical damping parameter $\gamma$?
   Answer: $\frac{\sqrt{3}}{2}$ For critical damping, $\gamma = \omega_0$. When damping is reduced by half, the new damping parameter becomes $\gamma_{new} = \gamma/2 = \omega_0/2$. For underdamped motion, the oscillation frequency is $\omega_d = \sqrt{\omega_0^2 - \gamma_{new}^2} = \sqrt{\omega_0^2 - (\omega_0/2)^2} = \sqrt{\omega_0^2 - \omega_0^2/4} = \sqrt{3\omega_0^2/4} = \frac{\sqrt{3}\omega_0}{2}$. Since $\omega_0 = \gamma$, we have $\omega_d = \frac{\sqrt{3}\gamma}{2}$. The ratio is $\frac{\omega_d}{\gamma} = \frac{\sqrt{3}}{2}$. Choice B incorrectly uses $\gamma/2$ in the denominator. Choice C ignores the frequency calculation. Choice D inverts the fraction.
2. The motion of a mass on a spring is described by the initial value problem $y'' + 16y = 0$, with $y(0) = \sqrt{3}$ and $y'(0) = 4$. If the solution is written in the form $y(t) = A \cos(4t - \phi)$, what is the phase shift $\phi$?
   Answer: $\phi = \pi/6$ The general solution to $y'' + 16y = 0$ is $y(t) = C_1 \cos(4t) + C_2 \sin(4t)$. The derivative is $y'(t) = -4C_1 \sin(4t) + 4C_2 \cos(4t)$. Using the initial conditions: $y(0) = C_1 = \sqrt{3}$ $y'(0) = 4C_2 = 4 \implies C_2 = 1$ So the solution is $y(t) = \sqrt{3} \cos(4t) + \sin(4t)$. To convert this to the form $y(t) = A \cos(4t - \phi)$, we use the relationships $C_1 = A \cos(\phi)$ and $C_2 = A \sin(\phi)$. The amplitude is $A = \sqrt{C_1^2 + C_2^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. The phase shift $\phi$ is determined by $\cos(\phi) = C_1/A = \sqrt{3}/2$ and $\sin(\phi) = C_2/A = 1/2$. Both conditions are satisfied by $\phi = \pi/6$.
3. A system is modeled by the differential equation $y'' + 100y = \cos(10.5t)$. The resulting motion exhibits a phenomenon known as beats, which is a periodic variation in amplitude. What is the frequency of these beats (in Hz)?
   Answer: $1/(4\pi)$ The phenomenon of beats arises from the superposition of two oscillations with close frequencies. The natural angular frequency of the system is $\omega_0 = \sqrt{100} = 10$ rad/s. The driving angular frequency is $\gamma = 10.5$ rad/s. The corresponding frequencies in Hz are $f_0 = \omega_0/(2\pi) = 10/(2\pi)$ and $f_d = \gamma/(2\pi) = 10.5/(2\pi)$. The beat frequency is the absolute difference between these two frequencies: $f_{beat} = |f_0 - f_d| = |10/(2\pi) - 10.5/(2\pi)| = | -0.5/(2\pi) | = 0.5/(2\pi) = 1/(4\pi)$ Hz.
4. The solution to an initial value problem $mx''+kx=0, x(0)=x_0, x'(0)=v_0$ is $x(t) = A\cos(\omega t - \phi)$. If the mass $m$ is quadrupled while $k$, $x_0$, and $v_0$ remain constant, how do the new angular frequency $\omega_{new}$ and amplitude $A_{new}$ relate to the original values?
   Answer: $\omega_{new} = \omega/2$ and $A_{new} = \sqrt{x_0^2 + 4v_0^2/\omega^2}$ The angular frequency is given by $\omega = \sqrt{k/m}$. If the mass $m$ is replaced by $m_{new} = 4m$, the new angular frequency is $\omega_{new} = \sqrt{k/(4m)} = (1/2)\sqrt{k/m} = \omega/2$. The solution is $x(t) = C_1\cos(\omega t) + C_2\sin(\omega t)$. The initial conditions give $C_1 = x(0) = x_0$ and $\omega C_2 = x'(0) = v_0$, so $C_2 = v_0/\omega$. The amplitude is $A = \sqrt{C_1^2 + C_2^2} = \sqrt{x_0^2 + (v_0/\omega)^2}$. For the new system, $C_{1,new} = x_0$ and $C_{2,new} = v_0/\omega_{new} = v_0/(\omega/2) = 2v_0/\omega$. The new amplitude is $A_{new} = \sqrt{C_{1,new}^2 + C_{2,new}^2} = \sqrt{x_0^2 + (2v_0/\omega)^2} = \sqrt{x_0^2 + 4v_0^2/\omega^2}$. This expression is not a simple multiple of the original amplitude $A$ unless $x_0=0$ or $v_0=0$.
5. Two identical oscillators with natural frequency $\omega_0$ are coupled, resulting in normal modes with frequencies $\omega_1 = 0.9\omega_0$ and $\omega_2 = 1.1\omega_0$. If the system is initiated such that both oscillators start from rest with equal displacements, and the motion can be written as a superposition of both modes with equal amplitudes, what is the beat frequency observed in either oscillator?
   Answer: $\frac{0.1\omega_0}{\pi}$ The motion of each oscillator can be written as a superposition: $x(t) = A\cos(\omega_1 t) + A\cos(\omega_2 t) = A\cos(0.9\omega_0 t) + A\cos(1.1\omega_0 t)$. Using the trigonometric identity $\cos(a) + \cos(b) = 2\cos(\frac{a+b}{2})\cos(\frac{a-b}{2})$, we get $x(t) = 2A\cos(\frac{0.9\omega_0 + 1.1\omega_0}{2}t)\cos(\frac{1.1\omega_0 - 0.9\omega_0}{2}t) = 2A\cos(\omega_0 t)\cos(0.1\omega_0 t)$. The beat frequency is the frequency of the amplitude modulation, which corresponds to $\cos(0.1\omega_0 t)$. However, since the amplitude goes through two maxima per period of the cosine function, the beat frequency is $f_{beat} = \frac{2 \times 0.1\omega_0}{2\pi} = \frac{0.1\omega_0}{\pi}$. Choice A uses the wrong coefficient. Choice C gives the angular frequency difference. Choice D gives half the angular frequency difference.
6. Consider the equation $y'' + \omega_0^2 y = F_0 \cos(\gamma t)$, whose steady-state solution has an amplitude $A(\gamma) = \frac{F_0}{|\omega_0^2 - \gamma^2|}$. A system with natural frequency $\omega_0=5$ rad/s is driven by a force with frequency $\gamma=4$ rad/s, producing a steady-state amplitude of 10 units. Which of the following driving frequencies $\gamma$ would produce a steady-state amplitude of 5 units?
   Answer: $\gamma = \sqrt{7}$ First, use the given information to find the magnitude of the driving force, $F_0$. We have $\omega_0=5$, $\gamma=4$, and $A(4)=10$. $A(4) = \frac{F_0}{|5^2 - 4^2|} = \frac{F_0}{|25 - 16|} = \frac{F_0}{9}$. Given $A(4)=10$, we have $10 = F_0/9$, so $F_0 = 90$. Next, we need to find $\gamma$ such that $A(\gamma)=5$. We use the value of $F_0$ we just found: $5 = \frac{90}{|5^2 - \gamma^2|} = \frac{90}{|25 - \gamma^2|}$. This implies $|25 - \gamma^2| = 90/5 = 18$. This gives two possibilities: 1) $25 - \gamma^2 = 18 \implies \gamma^2 = 7 \implies \gamma = \sqrt{7}$. 2) $25 - \gamma^2 = -18 \implies \gamma^2 = 43 \implies \gamma = \sqrt{43}$. Of the given answer choices, only $\gamma = \sqrt{7}$ is listed.
7. Two solutions to the differential equation $y'' + 4y = 0$, denoted $y_1(t)$ and $y_2(t)$, have the same amplitude. The graph of $y_2(t)$ is obtained by shifting the graph of $y_1(t)$ to the right by $\pi/8$ time units. If the solutions are written as $y_1(t) = A \cos(2t - \phi_1)$ and $y_2(t) = A \cos(2t - \phi_2)$, what is the value of the phase difference $\phi_2 - \phi_1$?
   Answer: $\pi/4$ From the differential equation $y'' + 4y = 0$, the angular frequency is $\omega = \sqrt{4} = 2$. A rightward time shift of $\Delta t$ corresponds to replacing $t$ with $t - \Delta t$. Thus, $y_2(t) = y_1(t - \Delta t)$. Substituting the forms of the solutions: $A \cos(2t - \phi_2) = A \cos(2(t - \Delta t) - \phi_1) = A \cos(2t - 2\Delta t - \phi_1)$. For the two expressions to be identical, their phases must be equal (or differ by a multiple of $2\pi$). So, $\phi_2 = \phi_1 + 2\Delta t$. The phase difference is $\phi_2 - \phi_1 = 2\Delta t$. Given the time shift $\Delta t = \pi/8$, the phase difference is $\phi_2 - \phi_1 = 2(\pi/8) = \pi/4$.
8. The motion of an underdamped oscillator is given by $y(t) = e^{-t}(3\cos(4t) - 3\sin(4t))$. The oscillatory part of the motion can be written as $A\cos(4t - \phi)$. What are the quasi-period $T_d$ of the motion and the phase shift $\phi$ in the interval $(-\pi, \pi]$?
   Answer: $T_d = \pi/2$, $\phi = -\pi/4$ The equation is of the form $y(t) = e^{-\alpha t} (C_1 \cos(\omega_d t) + C_2 \sin(\omega_d t))$. The quasi-frequency is $\omega_d = 4$ rad/s. The quasi-period is $T_d = 2\pi / \omega_d = 2\pi / 4 = \pi/2$. The oscillatory part is $3\cos(4t) - 3\sin(4t)$. Here, $C_1 = 3$ and $C_2 = -3$. The amplitude of this part is $A = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$. The phase shift $\phi$ is found from $\cos(\phi) = C_1/A = 3/(3\sqrt{2}) = 1/\sqrt{2}$ and $\sin(\phi) = C_2/A = -3/(3\sqrt{2}) = -1/\sqrt{2}$. The angle $\phi$ that satisfies both conditions (positive cosine, negative sine) is in the fourth quadrant. In the interval $(-\pi, \pi]$, this value is $\phi = -\pi/4$.
9. A damped harmonic oscillator satisfies the differential equation $m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$, where $m = 2$, $c = 4$, and $k = 10$. If the initial conditions are $x(0) = 3$ and $x'(0) = -1$, what is the amplitude of the oscillatory motion?
   Answer: $\sqrt{10}$ The characteristic equation is $2r^2 + 4r + 10 = 0$, giving $r^2 + 2r + 5 = 0$. Using the quadratic formula: $r = \frac{-2 \pm \sqrt{4-20}}{2} = -1 \pm 2i$. The general solution is $x(t) = e^{-t}(A\cos(2t) + B\sin(2t))$. From $x(0) = 3$: $A = 3$. From $x'(0) = -1$: $x'(t) = -e^{-t}(A\cos(2t) + B\sin(2t)) + e^{-t}(-2A\sin(2t) + 2B\cos(2t))$, so $x'(0) = -A + 2B = -3 + 2B = -1$, giving $B = 1$. The amplitude is $\sqrt{A^2 + B^2} = \sqrt{9 + 1} = \sqrt{10}$. Choice B incorrectly uses just the initial displacement. Choice C incorrectly adds the squares of initial conditions. Choice D incorrectly divides by the mass.
10. The equation of motion for a mass-spring system is $2x'' + kx = 0$. The period of oscillation is observed to be $\pi$ seconds. What is the value of the spring constant $k$?
    Answer: $k = 8$ First, we write the differential equation in standard form: $x'' + (k/2)x = 0$. The square of the angular frequency, $\omega^2$, is the coefficient of the $x$ term. So, $\omega^2 = k/2$, which means $\omega = \sqrt{k/2}$. The relationship between the period $T$ and the angular frequency $\omega$ is $T = 2\pi/\omega$. We are given that $T = \pi$. So, $\pi = 2\pi/\omega$, which implies $\omega = 2$ rad/s. Now we can find $k$: $\omega^2 = 4 = k/2$. Solving for $k$ gives $k = 8$.
11. An RLC circuit is governed by $L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0$, where $q(t)$ is the charge. If $L = 1$ H, $C = \frac{1}{25}$ F, and the circuit exhibits oscillatory behavior with a frequency that is $80\%$ of the natural frequency, what is the resistance $R$?
    Answer: $6$ Ω When you encounter RLC circuit problems, you're dealing with second-order linear differential equations that model oscillatory systems. The key insight is understanding how damping affects the natural frequency of oscillation. The given equation $L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0$ has characteristic equation $Ls^2 + Rs + \frac{1}{C} = 0$. For oscillatory behavior, we need complex roots, which occurs when the discriminant is negative: $R^2 < \frac{4L}{C}$. The natural (undamped) frequency is $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \cdot \frac{1}{25}}} = 5$ rad/s. The damped frequency is $\omega_d = \sqrt{\omega_0^2 - \left(\frac{R}{2L}\right)^2} = \sqrt{25 - \frac{R^2}{4}}$. Since the damped frequency is 80% of the natural frequency: $\omega_d = 0.8 \times 5 = 4$ rad/s. Setting up the equation: $4 = \sqrt{25 - \frac{R^2}{4}}$ Squaring both sides: $16 = 25 - \frac{R^2}{4}$ Solving: $\frac{R^2}{4} = 9$, so $R^2 = 36$, therefore $R = 6$ Ω. Choice (A) 10 Ω would make $R^2 = 100 > \frac{4L}{C} = 100$, creating the critically damped case with no oscillation. Choice (B) 8 Ω gives a damped frequency of 3 rad/s (60% of natural). Choice (C) 4 Ω yields a damped frequency of $\sqrt{21} \approx 4.58$ rad/s (about 92% of natural). Remember: in damped oscillators, always check that your resistance value allows oscillatory behavior by ensuring $R < 2\sqrt{\frac{L}{C}}$.
12. A solution to the differential equation $y'' + \omega^2 y = 0$ has an amplitude of 5, passes through the origin at $t=0$ with positive velocity, and reaches its first maximum displacement at $t = \pi/6$. What is the value of the angular frequency $\omega$?
    Answer: $\omega = 3$ The general solution is $y(t) = C_1\cos(\omega t) + C_2\sin(\omega t)$. The condition $y(0)=0$ implies $C_1\cos(0) + C_2\sin(0) = C_1 = 0$. So the solution is of the form $y(t) = C_2\sin(\omega t)$. The amplitude is $|C_2|$, so $|C_2|=5$. The velocity is $y'(t) = \omega C_2 \cos(\omega t)$. The condition $y'(0) > 0$ means $\omega C_2 > 0$. Since $\omega > 0$, we must have $C_2 > 0$, so $C_2=5$. The solution is $y(t) = 5\sin(\omega t)$. The maximum displacement is 5. The first time $t>0$ this occurs is when $\sin(\omega t) = 1$. The smallest positive value for the argument is $\omega t = \pi/2$. We are given that this occurs at $t=\pi/6$. Therefore, $\omega(\pi/6) = \pi/2$. Solving for $\omega$ gives $\omega = (\pi/2) / (\pi/6) = 3$.
13. The position of a mass oscillating on a spring is given by $x(t) = \cos(2t) + \sqrt{3} \sin(2t)$. What is the first time $t > 0$ that the mass reaches its maximum positive displacement from equilibrium?
    Answer: $t = \pi/6$ First, we convert the solution to amplitude-phase form, $x(t) = A \cos(\omega t - \phi)$. The angular frequency is $\omega=2$. The coefficients are $C_1=1$ and $C_2=\sqrt{3}$. The amplitude is $A = \sqrt{C_1^2 + C_2^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$. The maximum displacement is $2$. The phase shift $\phi$ satisfies $\cos(\phi) = C_1/A = 1/2$ and $\sin(\phi) = C_2/A = \sqrt{3}/2$. This means $\phi = \pi/3$. So, $x(t) = 2\cos(2t - \pi/3)$. The mass reaches its maximum displacement when $x(t)=2$, which means $\cos(2t - \pi/3) = 1$. This occurs when the argument of the cosine is $2n\pi$ for any integer $n$. We seek the smallest $t>0$. $2t - \pi/3 = 2n\pi$. For $n=0$, we have $2t - \pi/3 = 0$, which gives $t = \pi/6$. This is the first positive time.
14. A pendulum undergoes small oscillations described by $\theta(t) = \theta_0 e^{-\beta t}\cos(\omega t + \phi)$. If the amplitude decreases to $\frac{1}{e^2}$ of its initial value in time $T$, and during this same time the pendulum completes exactly $3.5$ oscillations, what is the ratio $\frac{\omega}{\beta}$?
    Answer: $\frac{7\pi}{2}$ When you encounter damped oscillation problems, you need to extract information from both the exponential decay (damping) and the oscillatory motion separately. The given solution $\theta(t) = \theta_0 e^{-\beta t}\cos(\omega t + \phi)$ shows that the amplitude is $\theta_0 e^{-\beta t}$. If this decreases to $\frac{1}{e^2}$ of its initial value in time $T$, then: $\theta_0 e^{-\beta T} = \frac{\theta_0}{e^2}$ This simplifies to $e^{-\beta T} = e^{-2}$, so $\beta T = 2$. For the oscillatory part, completing 3.5 oscillations in time $T$ means the phase $\omega T$ advances by $3.5 \times 2\pi = 7\pi$ radians. Therefore: $\omega T = 7\pi$. Now you can find the ratio: $\frac{\omega}{\beta} = \frac{\omega T}{\beta T} = \frac{7\pi}{2}$ Looking at the wrong answers: Choice A ($\frac{7\pi}{4}$) results from mistakenly using 1.75 oscillations instead of 3.5. Choice B ($7\pi$) comes from setting $\beta T = 1$ instead of 2, likely confusing the amplitude reduction factor. Choice D ($14\pi$) appears when you incorrectly double the oscillation count or mishandle the exponential decay relationship. The correct answer is C. Study tip: In damped oscillation problems, always handle the exponential and trigonometric parts separately. The exponential gives you damping information, while the argument of the cosine gives you frequency information. Set up your equations independently, then combine the results.

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