Differential Equations Quiz: Amplitude Frequency And Phase

What this quiz covers

This quiz focuses on Amplitude Frequency And Phase, 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. $\frac{\sqrt{3}}{2}$ (correct answer)
2. $\frac{\sqrt{3}}{4}$
3. $\frac{1}{2}$
4. $\sqrt{3}$

Explanation: 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.

Question 2

1. $\phi = \pi/6$ (correct answer)
2. $\phi = \pi/3$
3. $\phi = \pi/4$
4. $\phi = \pi/2$

Explanation: 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 $\phi = \pi/6$.

Question 3

1. $\pi/2$
2. $\pi/4$ (correct answer)
3. $\pi/8$
4. $-\pi/4$

Explanation: 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 $\phi_2 - \phi_1 = 2(\pi/8) = \pi/4$.

Question 4

1. $t = \pi/12$
2. $t = \pi/6$ (correct answer)
3. $t = 5\pi/12$
4. $t = \pi/3$

Explanation: 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.

Question 5

1. $k = 2$
2. $k = 4$
3. $k = 8$ (correct answer)
4. $k = 16$

Explanation: 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 $k = 8$.

Question 6

1. $T_d = \pi/2$, $\phi = \pi/4$
2. $T_d = 2\pi$, $\phi = -\pi/4$
3. $T_d = \pi/2$, $\phi = -\pi/4$ (correct answer)
4. $T_d = \pi/2$, $\phi = 3\pi/4$

Explanation: 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 $\phi = -\pi/4$.

Question 7

1. $\omega = 1/3$
2. $\omega = 3/2$
3. $\omega = 3$ (correct answer)
4. $\omega = 6$

Explanation: 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 $\omega = (\pi/2) / (\pi/6) = 3$.

Question 8

1. $\omega_{new} = \omega/2$ and $A_{new} = \sqrt{x_0^2 + 4v_0^2/\omega^2}$ (correct answer)
2. $\omega_{new} = \omega/4$ and $A_{new} = A/4$
3. $\omega_{new} = 2\omega$ and $A_{new} = \sqrt{x_0^2 + v_0^2/(4\omega^2)}$
4. $\omega_{new} = \omega/2$ and $A_{new} = A/2$

Explanation: 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 $v_0=0$.

Question 9

1. $\gamma = \sqrt{7}$ (correct answer)
2. $\gamma = 3$
3. $\gamma = \sqrt{17}$
4. $\gamma = 2$

Explanation: 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.

Question 10

1. $0.5$
2. $1/(4\pi)$ (correct answer)
3. $1/(8\pi)$
4. $20.5/(2\pi)$

Explanation: 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.

Question 11

1. $\frac{0.2\omega_0}{2\pi}$
2. $\frac{0.1\omega_0}{\pi}$ (correct answer)
3. $0.2\omega_0$
4. $0.1\omega_0$

Explanation: 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.

Question 12

1. $\sqrt{10}$ (correct answer)
2. $3$
3. $\sqrt{13}$
4. $\frac{\sqrt{10}}{2}$

Explanation: 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.

Question 13

1. $\frac{7\pi}{4}$
2. $7\pi$
3. $\frac{7\pi}{2}$ (correct answer)
4. $14\pi$

Explanation: 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.

Question 14

1. $10$ Ω
2. $8$ Ω
3. $4$ Ω
4. $6$ Ω (correct answer)

Explanation: 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 $R < 2\sqrt{\frac{L}{C}}$.