Statics and Dynamics Quiz: Undamped Free Vibration

What this quiz covers

This quiz focuses on Undamped Free Vibration, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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. $\omega_n = \sqrt{\dfrac{3g}{2L}}$, derived by treating the rod as a physical pendulum with moment of inertia $\frac{1}{3}mL^2$ about the pin and restoring moment arm $\frac{L}{2}$. (correct answer)
2. $\omega_n = \sqrt{\dfrac{g}{L}}$, derived by treating the system as a simple (point-mass) pendulum of length $L$ with the full rod length used as the effective pendulum length.
3. $\omega_n = \sqrt{\dfrac{2g}{L}}$, derived by using the moment of inertia $\frac{1}{2}mL^2$ about the pin, which applies to a thin disk rather than a slender rod rotating about its end.
4. $\omega_n = \sqrt{\dfrac{g}{2L}}$, derived by treating the center of mass as the pivot point with moment of inertia $\frac{1}{12}mL^2$ and restoring moment arm equal to the full rod length $L$.

Question 2

1. The mass must be multiplied by a factor of 2, since natural frequency is inversely proportional to the square root of mass, so halving the frequency requires doubling the mass: $\omega_n \propto 1/\sqrt{M}$ implies $M_{new} = 4M_{old}$... but only a factor of 2 is needed for a 50% change in $\sqrt{M}$.
2. The mass must be multiplied by a factor of 4, since $\omega_n = \sqrt{k/M}$ and halving $\omega_n$ requires $\omega_{n,new}^2 = k/M_{new} = (\omega_{n,old}/2)^2 = \omega_{n,old}^2/4$, which gives $M_{new} = 4M_{old}$. (correct answer)
3. The mass must be multiplied by a factor of $\sqrt{2}$, since a 50% reduction in frequency corresponds to a 50% reduction in $\omega_n^2$, and $M_{new}/M_{old} = \omega_{n,old}^2/\omega_{n,new}^2 = 1/0.5 = \sqrt{2}$ after taking the square root of the frequency ratio.
4. The mass must be multiplied by a factor of 2, since $\omega_n \propto 1/\sqrt{M}$ and a 50% reduction in frequency means $\omega_{n,new}/\omega_{n,old} = 0.5$, which directly gives $M_{new}/M_{old} = 1/0.5 = 2$ by simple proportionality without squaring.

Question 3

1. $v_{max} = 0.40\text{ m/s}$, computed by differentiating the sine term only, since $\dot{x} = -0.03(10)\sin(10t) + 0.04(10)\cos(10t)$ and the maximum of the cosine term $0.04 \times 10 = 0.40\text{ m/s}$ dominates at $t = 0$.
2. $v_{max} = 0.70\text{ m/s}$, computed as $\omega_n$ times the sum of the individual amplitudes $(0.03 + 0.04) = 0.07\text{ m}$, since both cosine and sine components reach their maxima and the speeds add directly at the instant of maximum velocity.
3. $v_{max} = 0.50\text{ m/s}$, computed as $\omega_n \cdot A$ where the amplitude $A = \sqrt{(0.03)^2 + (0.04)^2} = 0.05\text{ m}$ and $\omega_n = 10\text{ rad/s}$, giving $10 \times 0.05 = 0.50\text{ m/s}$. (correct answer)
4. $v_{max} = 0.30\text{ m/s}$, computed as $\omega_n$ times the cosine amplitude alone $(0.03\text{ m})$, since the cosine term represents the displacement-driven component and the sine term represents the velocity-driven component which does not contribute to the speed envelope.

Question 4

1. The solution is fully correct, because the amplitude of oscillation is $v_0/\omega_n$ regardless of the sign of the initial velocity, and the direction of the first quarter-cycle does not affect the amplitude expression for undamped free vibration.
2. The solution is correct in form but uses the wrong frequency, because for a vertical spring-mass system the effective natural frequency includes the gravitational acceleration: $\omega_{n,eff} = \sqrt{(k + mg)/m}$, and the corrected solution replaces $\omega_n$ with $\omega_{n,eff}$.
3. The solution is incorrect in both sign and form, because measuring from static equilibrium in the vertical direction requires gravity to appear explicitly in the equation of motion, adding a particular solution term $\frac{g}{\omega_n^2}(1 - \cos(\omega_n t))$ to the response.
4. The solution has an incorrect sign. With $x$ positive downward, an initial upward velocity means $\dot{x}(0) = -v_0$. The correct solution is $x(t) = -\frac{v_0}{\omega_n}\sin(\omega_n t)$, which has the same amplitude but correctly predicts that the mass initially moves upward (negative $x$). (correct answer)

Question 5

1. $\omega_n = \sqrt{\dfrac{k}{m}}$, since the spring force acts at the center of mass and the rolling constraint does not affect the translational equation of motion when the contact point is frictionless.
2. $\omega_n = \sqrt{\dfrac{2k}{3m}}$, since rolling without slipping introduces a rotational kinetic energy term that increases the effective inertia. The disk's moment of inertia $\frac{1}{2}mR^2$ and rolling constraint $\dot{\theta} = \dot{x}/R$ yield an effective mass of $\frac{3}{2}m$, giving $\omega_n = \sqrt{k/(3m/2)}$. (correct answer)
3. $\omega_n = \sqrt{\dfrac{k}{2m}}$, since the rolling constraint causes the effective inertia to equal $2m$, because the disk's moment of inertia about the contact point is $\frac{3}{2}mR^2$ and the rotational equation about the contact point replaces the translational one, doubling the apparent mass.
4. $\omega_n = \sqrt{\dfrac{2k}{m}}$, since the no-slip condition at the contact point means the spring is effectively twice as stiff — the contact point is momentarily fixed, so the disk behaves as if attached to a spring of stiffness $2k$ while its full mass $m$ provides the only inertia.

Question 6

1. The natural frequency decreases by a factor of 2, since the series equivalent stiffness $k_{eq} = k/2$ is one-quarter of the parallel equivalent stiffness $k_{eq} = 2k$, and frequency scales as the square root of stiffness, giving $\sqrt{1/4} = 1/2$. (correct answer)
2. The natural frequency decreases by a factor of 4, since the equivalent stiffness changes from $2k$ (parallel) to $k/2$ (series), and natural frequency is directly proportional to equivalent stiffness, so the ratio is $k/2 \div 2k = 1/4$.
3. The natural frequency decreases by a factor of $\sqrt{2}$, since the series stiffness $k/2$ is half the individual spring stiffness $k$, and comparing only to the single-spring baseline gives a factor of $1/\sqrt{2}$, ignoring the parallel configuration's effect.
4. The natural frequency decreases by a factor of $\sqrt{2}$, since going from parallel to series halves the equivalent stiffness from $2k$ to $k$, and a halving of stiffness reduces frequency by $\sqrt{1/2} = 1/\sqrt{2}$, because the series stiffness is mistakenly taken as $k$ rather than $k/2$.

Question 7

1. $A = 0.05\text{ m}$, since only the initial displacement contributes to amplitude when the initial velocity is in the direction that increases displacement, and the velocity term cancels in the amplitude formula.
2. $A = 0.10\text{ m}$, since the amplitude is the algebraic sum of the initial displacement and the ratio of initial velocity to natural frequency, giving $0.05 + 0.05 = 0.10\text{ m}$.
3. $A \approx 0.0707\text{ m}$, since the amplitude is found from $A = \sqrt{x_0^2 + (v_0/\omega_n)^2}$ with $\omega_n = 10\text{ rad/s}$, yielding $\sqrt{(0.05)^2 + (0.10)^2} \approx 0.1118\text{ m}$, which rounds to $0.0707\text{ m}$ after adjusting units.
4. $A \approx 0.1118\text{ m}$, since the amplitude is $A = \sqrt{x_0^2 + (v_0/\omega_n)^2}$ with $\omega_n = \sqrt{k/m} = 10\text{ rad/s}$, giving $\sqrt{(0.05)^2 + (1/10)^2} = \sqrt{0.0025 + 0.01} = \sqrt{0.0125}\text{ m}$. (correct answer)

Question 8

1. The natural frequency is $\omega_n = \sqrt{k\cos\theta/m}$, because only the component of spring force perpendicular to the gravitational vector contributes to the restoring mechanism, and the full stiffness $k$ must be projected by $\cos\theta$.
2. The natural frequency is $\omega_n = \sqrt{(k + mg\sin\theta)/m}$, because the gravitational component along the incline acts as an additional restoring force that adds to the spring stiffness and increases the effective stiffness of the system.
3. The natural frequency is $\omega_n = \sqrt{k/m}$, and gravity does not appear in the equation of motion for small oscillations because the gravitational component $mg\sin\theta$ along the incline is a constant force that shifts the equilibrium position but cancels out when the equation is written in terms of displacement from static equilibrium. (correct answer)
4. The natural frequency is $\omega_n = \sqrt{(k - mg\sin\theta)/m}$, because gravity acts opposite to the spring force along the incline and reduces the effective restoring stiffness, causing the system to vibrate more slowly than a horizontal spring-mass system.

Question 9

1. The period of System II is 4 times that of System I, and the maximum kinetic energy of System II is equal to that of System I, since both systems are given the same initial displacement and total mechanical energy is identical regardless of spring stiffness.
2. The period of System II is 4 times that of System I, and the maximum kinetic energy of System II is 1/4 times that of System I, since the softer spring in System II stores less initial potential energy $\frac{1}{2}(k/4)x_0^2$ at the same displacement, and all of that energy converts to kinetic energy at equilibrium. (correct answer)
3. The period of System II is 16 times that of System I, and the maximum kinetic energy of System II is 1/4 times that of System I, since the period scales with the ratio $m/k$ directly (not its square root) and potential energy scales linearly with stiffness.
4. The period of System II is 4 times that of System I, and the maximum kinetic energy of System II is 4 times that of System I, since the heavier mass of System II reaches greater kinetic energy at the equilibrium position due to greater inertia driving the motion.