Undamped Free Vibration - Statics and Dynamics

1. If only the mass $M$ can be modified (the column stiffness is fixed), by what factor must $M$ be multiplied to achieve the desired reduction in natural frequency?
   Answer: 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 $M_{new} = 4M_{old}$. Whenever you see a vibration question asking how a parameter change affects natural frequency, your first instinct should be to write the governing formula and manipulate it algebraically — never reason by "direct proportionality" without tracking the exponents carefully. For a lumped-mass system, the natural frequency is $\omega_n = \sqrt{k/M}$. If you want $\omega_{n,new} = 0.5\,\omega_{n,old}$, square both sides of the ratio: $\frac{\omega_{n,new}^2}{\omega_{n,old}^2} = \frac{k/M_{new}}{k/M_{old}} = \frac{M_{old}}{M_{new}} = (0.5)^2 = 0.25$ Solving gives $M_{new} = 4\,M_{old}$, a factor of 4. That confirms B is correct. Choice A starts with the right formula but then contradicts itself — it states the factor is 2 while simultaneously deriving $M_{new} = 4M_{old}$. The internal inconsistency should be a red flag; always trust the algebra over a verbal shortcut. Choice C makes a critical error by claiming a 50% reduction in $\omega_n$ means a 50% reduction in $\omega_n^2$. In reality, $(0.5)^2 = 0.25$, not 0.5 — squaring the frequency ratio is essential, and this distractor skips that step entirely. Choice D treats the relationship as linear: $\omega_n \propto 1/M$ instead of $\omega_n \propto 1/\sqrt{M}$. Dropping the square root is one of the most common algebra errors in dynamics problems and always leads to an answer that is too small. Study tip: Any time you adjust a parameter to hit a target frequency, square the frequency ratio first, then solve for the unknown. The squaring step is where most mistakes happen.
2. Which expression correctly gives the natural frequency of small oscillation for this system?
   Answer: $\omega_n = \sqrt{\dfrac{3g}{2L}}$, 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 $\frac{L}{2}$. When a rigid body swings about a fixed pivot, you're dealing with a physical pendulum, not a simple one. The key equation is the rotational form of Newton's second law: the net restoring torque equals the moment of inertia times angular acceleration. For small angles, this gives $\omega_n = \sqrt{\frac{mgd}{I_O}}$, where $d$ is the distance from the pivot to the center of mass, and $I_O$ is the moment of inertia about the pivot. For a uniform slender rod pinned at one end, the center of mass sits at $L/2$, so the restoring torque arm is $d = L/2$. The moment of inertia about the pin (using the parallel axis theorem or standard result) is $I_O = \frac{1}{3}mL^2$. Plugging in: $\omega_n = \sqrt{\frac{mg(L/2)}{\frac{1}{3}mL^2}} = \sqrt{\frac{3g}{2L}}$. That's exactly answer A, confirming it as correct. Answer B applies the simple pendulum formula $\sqrt{g/L}$, which assumes all mass is concentrated at the tip — ignoring how the rod's distributed mass actually behaves rotationally. Answer C uses $I = \frac{1}{2}mL^2$, which is the moment of inertia for a thin disk about its central axis — a completely wrong geometry. Answer D incorrectly places the pivot at the center of mass; you can't generate a gravitational restoring torque about the center of mass itself, and mixing $\frac{1}{12}mL^2$ with a full-length moment arm is physically inconsistent. Your go-to study tip: whenever you see a rigid body swinging about a pin, immediately write down $I_O$ and $d$ before doing anything else — getting those two quantities right is the entire problem.
3. How does the period of System II compare to that of System I, and how does the maximum kinetic energy of System II compare to that of System I?
   Answer: 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. When you encounter a spring-mass vibration problem, anchor yourself to two formulas: the natural period $T = 2\pi\sqrt{m/k}$ and the initial potential energy $PE = \frac{1}{2}kx_0^2$. These two expressions govern everything here. For System II, the period becomes $T_{II} = 2\pi\sqrt{4m/(k/4)} = 2\pi\sqrt{16m/k} = 4 \cdot 2\pi\sqrt{m/k} = 4T_I$. The period quadruples because you're taking the square root of a ratio that increased by a factor of 16. Now for energy: the initial potential energy stored in System I is $\frac{1}{2}kx_0^2$, while System II stores $\frac{1}{2}(k/4)x_0^2 = \frac{1}{4} \cdot \frac{1}{2}kx_0^2$. Since both systems are undamped, all initial potential energy converts to kinetic energy at equilibrium. System II therefore reaches only 1/4 the maximum kinetic energy of System I. That confirms B is correct. Choice A fails on energy: it wrongly assumes both systems store identical potential energy at the same displacement, ignoring that softer springs store less energy ($PE \propto k$). Choice C gets the energy comparison right but makes a critical error on the period, claiming it scales with $m/k$ directly rather than $\sqrt{m/k}$—that ratio is 16, but the period only quadruples after taking the square root. Choice D inverts the energy logic; greater inertia does not generate extra energy—energy is set entirely by the initial conditions and stiffness, not mass alone. A useful habit: always write out $T = 2\pi\sqrt{m/k}$ explicitly before computing ratios, and track potential energy through $\frac{1}{2}kx_0^2$ rather than intuition about "heavier" systems.
4. If the two springs are then rearranged into a series configuration (one spring connects ceiling to an intermediate point, and the second spring connects that point to the mass), by what factor does the natural frequency change?
   Answer: 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 $\sqrt{1/4} = 1/2$. When you encounter spring combination problems, always work through two distinct steps: find the equivalent stiffness for each configuration, then apply the natural frequency formula $\omega_n = \sqrt{k_{eq}/m}$. The key insight is that frequency scales as the square root of stiffness, not linearly. Start with the two configurations. Springs in parallel add directly: $k_{parallel} = k + k = 2k$. Springs in series combine as reciprocals: $\frac{1}{k_{series}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$, giving $k_{series} = k/2$. The ratio of stiffnesses is $\frac{k/2}{2k} = \frac{1}{4}$. Since $\omega_n \propto \sqrt{k_{eq}}$, the frequency ratio is $\sqrt{1/4} = \frac{1}{2}$. The natural frequency is cut in half — confirming answer A is correct. Answer B makes the classic mistake of treating natural frequency as directly proportional to stiffness rather than proportional to its square root. The stiffness ratio of $1/4$ does not mean the frequency drops by a factor of 4. Answer C compares the series stiffness only to a single spring's stiffness $k$, completely ignoring the parallel configuration. You must compare $k_{series}$ to $k_{parallel}$, not to an arbitrary baseline. Answer D correctly identifies the frequency-stiffness relationship but miscalculates the series stiffness as $k$ instead of $k/2$, making the stiffness ratio $1/2$ rather than the correct $1/4$. Always recompute series stiffness carefully using the reciprocal formula. Study tip: Memorize the "square root rule" — whenever stiffness changes by a factor of $n$, frequency changes by $\sqrt{n}$. This catches both the B-type and D-type errors in one shot.
5. What is the maximum speed of the block during vibration?
   Answer: $v_{max} = 0.50\text{ m/s}$, 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 $10 \times 0.05 = 0.50\text{ m/s}$. When a vibration problem gives you a response like $x(t) = C_1\cos(\omega_n t) + C_2\sin(\omega_n t)$, your first move should be to recognize that this is a single harmonic oscillation written in two-component form. The true amplitude isn't found by adding or cherry-picking the individual coefficients — it's found using the Pythagorean combination: $A = \sqrt{C_1^2 + C_2^2}$. Once you have $A$, the maximum speed is simply $v_{max} = \omega_n A$, because velocity is $\dot{x}(t) = \omega_n(-C_1\sin(\omega_n t) + C_2\cos(\omega_n t))$, which is itself a sinusoid with amplitude $\omega_n\sqrt{C_1^2 + C_2^2}$. Here, $A = \sqrt{(0.03)^2 + (0.04)^2} = \sqrt{0.0009 + 0.0016} = \sqrt{0.0025} = 0.05\text{ m}$, and with $\omega_n = 10\text{ rad/s}$, you get $v_{max} = 10 \times 0.05 = 0.50\text{ m/s}$, confirming C. A is wrong because it evaluates the velocity expression only at $t = 0$, which happens to zero out the sine term — that's one specific instant, not the maximum. B adds the amplitudes linearly ($0.03 + 0.04 = 0.07$), which would only be valid if both components peaked simultaneously, but two sinusoids 90° out of phase never do — their combined peak follows the Pythagorean rule. D uses only the cosine coefficient and ignores the sine term entirely, discarding half the motion's energy with no physical justification. Study tip: Any time you see $A\cos(\omega t) + B\sin(\omega t)$, immediately think "Pythagorean amplitude" — $\sqrt{A^2 + B^2}$. This pattern appears everywhere in vibrations and AC circuits, and linearly adding or ignoring terms is one of the most common traps on dynamics exams.
6. What is the natural frequency of the resulting undamped free vibration, accounting for the rolling constraint?
   Answer: $\omega_n = \sqrt{\dfrac{2k}{3m}}$, 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 $\omega_n = \sqrt{k/(3m/2)}$. When a disk rolls without slipping, you must account for both translational and rotational kinetic energy — the rolling constraint couples them together, effectively increasing the system's inertia and lowering the natural frequency compared to a simple mass-spring system. To derive the equation of motion, apply the Lagrangian (or Newton's laws with the rolling constraint $\dot{\theta} = \dot{x}/R$). The total kinetic energy is: $T = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\dot{\theta}^2 = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\frac{\dot{x}^2}{R^2} = \frac{3}{4}m\dot{x}^2$ This gives an effective mass of $\frac{3}{2}m$. With potential energy $V = \frac{1}{2}kx^2$, the equation of motion becomes $\frac{3}{2}m\ddot{x} + kx = 0$, yielding: $\omega_n = \sqrt{\frac{k}{3m/2}} = \sqrt{\frac{2k}{3m}}$ This confirms B is correct. A is wrong because it ignores rotational inertia entirely — the rolling constraint absolutely affects dynamics by adding the $\frac{1}{2}I\dot{\theta}^2$ term, even if friction does no work. C is wrong due to an incorrect moment of inertia. The parallel-axis theorem gives $I_{contact} = \frac{3}{2}mR^2$, but using torques about the contact point still yields the same equation of motion as above — not an effective mass of $2m$. D is wrong because the no-slip condition constrains kinematics, not spring stiffness. The spring stretches by $x$, so its effective stiffness remains $k$. Study tip: Whenever you see "rolls without slipping," immediately write out the full kinetic energy including rotation. The effective mass for a solid disk is always $\frac{3}{2}m$ — memorize this result.
7. Which statement correctly describes the natural frequency of the resulting undamped vibration and the role of gravity in the equation of motion?
   Answer: 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. Whenever you see a spring-mass system on an inclined plane, your first instinct should be to ask: where is the equilibrium position, and what happens when I measure displacement from there? Start by writing Newton's second law along the incline. Let $x$ be displacement measured from the unstretched spring position. The spring force is $-kx$ (up the incline) and gravity contributes $-mg\sin\theta$ (down the incline), giving: $m\ddot{x} = -kx - mg\sin\theta$ Now shift to a new coordinate $u = x - x_{eq}$, where the static equilibrium point $x_{eq} = -mg\sin\theta / k$ satisfies the equilibrium condition. Substituting, the constant gravity term cancels exactly, leaving: $m\ddot{u} = -ku \implies \omega_n = \sqrt{k/m}$ This confirms C: gravity shifts the equilibrium position but vanishes from the dynamic equation, so the natural frequency is identical to a horizontal spring-mass system. A is wrong because there is no reason to project the spring stiffness by $\cos\theta$. The spring acts along the incline, so its full stiffness $k$ already operates in the direction of motion — no projection needed. B is wrong because gravity is a constant force, not a stiffness. It doesn't grow with displacement, so it cannot add to the effective spring constant. Only displacement-dependent forces alter $\omega_n$. D makes the same category error as B, treating gravity as if it subtracts from stiffness. Gravity has no $x$-dependence, so it contributes nothing to the restoring force dynamics. The key study tip: constant forces shift equilibrium; they never change natural frequency. This principle applies to gravity on any incline, pre-tension in cables, or buoyancy in fluid problems.
8. What is the amplitude of the resulting undamped free vibration?
   Answer: $A \approx 0.1118\text{ m}$, 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 $\sqrt{(0.05)^2 + (1/10)^2} = \sqrt{0.0025 + 0.01} = \sqrt{0.0125}\text{ m}$. Whenever you encounter undamped free vibration problems, your first instinct should be to find the natural frequency, then apply the amplitude formula — these two steps unlock everything else. Start with the natural frequency: $\omega_n = \sqrt{k/m} = \sqrt{400/4} = 10 \text{ rad/s}$. For undamped free vibration, the general solution is $x(t) = x_0\cos(\omega_n t) + \frac{v_0}{\omega_n}\sin(\omega_n t)$. The amplitude of a sum of a sine and cosine with the same frequency is found using the Pythagorean relationship — not simple addition — giving $A = \sqrt{x_0^2 + (v_0/\omega_n)^2}$. Plugging in: $A = \sqrt{(0.05)^2 + (1/10)^2} = \sqrt{0.0025 + 0.01} = \sqrt{0.0125} \approx 0.1118 \text{ m}$. That confirms D. A contains a fundamental misconception — the velocity term does not cancel regardless of its direction. Both initial conditions independently contribute to amplitude through the sine and cosine components respectively. B makes the critical error of adding $x_0$ and $v_0/\omega_n$ algebraically ($0.05 + 0.10 = 0.10$... note it even gets the arithmetic wrong, since $1/10 = 0.10$, making the sum $0.15$, not $0.10$). Amplitude comes from vector addition of orthogonal components, not scalar addition. C uses the correct formula but then bizarrely claims $\sqrt{(0.05)^2 + (0.10)^2} \approx 0.0707$, which is actually $\approx 0.1118$. The $0.0707$ figure comes from $\frac{0.1}{\sqrt{2}}$, suggesting a unit-conversion error was invented to manufacture a wrong answer. Remember: whenever two sinusoidal components are combined, amplitude is always the square root of the sum of squares, never the direct sum.
9. The student's solution correctly predicts the natural frequency but may contain an error in the initial condition application. Which statement is most accurate?
   Answer: 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 $x$). When a question gives you a coordinate system and an initial condition, your first job is to translate the physical description into math using that coordinate system — sign and all. Here, $x$ is defined as positive downward. An initial upward velocity is physically in the negative $x$-direction, so the correct initial condition is $\dot{x}(0) = -v_0$, not $+v_0$. For undamped free vibration starting from equilibrium, the general solution is $x(t) = A\sin(\omega_n t) + B\cos(\omega_n t)$. Applying $x(0) = 0$ gives $B = 0$, and applying $\dot{x}(0) = -v_0$ gives $A = -v_0/\omega_n$. The correct solution is therefore $x(t) = -\frac{v_0}{\omega_n}\sin(\omega_n t)$, confirming D. This predicts the mass initially moves in the negative $x$ direction (upward), which is physically sensible. A is wrong because the sign of the initial velocity absolutely matters for predicting the direction of motion. Amplitude is always positive, but the signed coefficient $A$ encodes which way the mass moves first — dropping that sign gives you a solution that's a mirror image in time of the true motion. B is wrong because gravity does not alter the natural frequency when you measure from the static equilibrium position. At static equilibrium, the spring is already pre-loaded by $mg/k$, so gravity cancels out of the equation of motion entirely. $\omega_n = \sqrt{k/m}$ remains unchanged. C is wrong for the same reason as B. When you measure from static equilibrium, there is no explicit gravity term or particular solution — the equation of motion is simply $\ddot{x} + \omega_n^2 x = 0$. Study tip: Always write out the sign of every initial condition explicitly before substituting. A single sign error here flips the entire trajectory.

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