Mass-Spring Systems - Statics and Dynamics

1. Which of the following correctly evaluates the colleague's suggestion?
   Answer: The colleague is incorrect. Increasing the mass by factor $n$ reduces the natural frequency by $\sqrt{n}$, giving $\omega_n' = \omega_n/\sqrt{n}$. Replacing the spring with stiffness $k/n^2$ gives $\omega_n'' = \sqrt{(k/n^2)/m} = \omega_n/n$. Since $\omega_n/n \neq \omega_n/\sqrt{n}$ for $n > 1$, the two strategies are not equivalent; to match the mass-addition strategy, the spring stiffness should be reduced to $k/n$, not $k/n^2$. When analyzing vibration isolation, the key formula is the natural frequency of a spring-mass system: $\omega_n = \sqrt{k/m}$. Whenever a question asks whether two design changes are "equivalent," your job is to compute the resulting frequency for each strategy separately and compare them directly — don't assume equivalence without checking. Start with the original frequency $\omega_n = \sqrt{k/m}$. If you increase mass by factor $n$, the new frequency is $\omega_n' = \sqrt{k/(nm)} = \omega_n/\sqrt{n}$. Now evaluate the colleague's spring strategy: replacing $k$ with $k/n^2$ gives $\omega_n'' = \sqrt{(k/n^2)/m} = \sqrt{k/m} \cdot (1/n) = \omega_n/n$. Since $\omega_n/n \neq \omega_n/\sqrt{n}$ for any $n > 1$, the strategies are not equivalent. To match the mass strategy, you'd need stiffness $k/n$ (not $k/n^2$), because $\sqrt{(k/n)/m} = \omega_n/\sqrt{n}$. This confirms C is correct. A makes a critical error in reasoning: yes, $\omega_n/n < \omega_n/\sqrt{n}$, but "overshooting" the target frequency doesn't make a design suggestion "acceptable" — the colleague claimed equivalence, which is simply false. B claims $\omega_n/\sqrt{n^2} = \omega_n/\sqrt{n}$, which is only true if $n = 1$ — it incorrectly equates $n$ with $\sqrt{n}$. D argues that $k/(nm) = (k/n^2)/m$, but simplifying both sides gives $k/(nm)$ vs. $k/(n^2 m)$ — these are only equal when $n = 1$. Your study tip: always substitute and simplify each scenario independently before comparing. Proportionality arguments like "both involve dividing $k$ and multiplying $m$" are frequent traps — the exponents matter enormously inside a square root.
2. What is the natural frequency of small oscillations of the pinned bar–spring system described above?
   Answer: $\omega_n = \sqrt{\dfrac{3k}{m}}$, because taking moments about the pin converts the translational spring stiffness into an effective rotational stiffness of $kL^2$ and the bar's moment of inertia about the pin is $\frac{1}{3}mL^2$, giving $\frac{kL^2}{mL^2/3}$. When a rigid bar rotates about a fixed pin, you should immediately shift your thinking from translational dynamics to rotational dynamics: apply Newton's second law for rotation, $\sum M = I\alpha$, rather than $F = ma$. For a small angular displacement $\theta$ from horizontal, the spring at the tip deflects by $L\theta$, producing a restoring force $kL\theta$. Taking moments about the pin, that force creates a restoring torque $kL^2\theta$. This is your effective rotational stiffness: $k_{\text{eff}} = kL^2$. The bar's moment of inertia about the pin is $I = \frac{1}{3}mL^2$. The equation of motion becomes $\frac{1}{3}mL^2\ddot{\theta} + kL^2\theta = 0$, giving: $\omega_n = \sqrt{\frac{k_{\text{eff}}}{I}} = \sqrt{\frac{kL^2}{\frac{1}{3}mL^2}} = \sqrt{\frac{3k}{m}}$ This confirms answer B is correct. A is wrong because it ignores the bar's rotational inertia entirely — you cannot treat this as a simple mass-spring system without accounting for how inertia is distributed along the bar. C inverts the logic: the effective mass at the tip decreases (to $m/3$, not $3m$), which would actually increase frequency, not decrease it — and the formula is still set up incorrectly. D is a tempting trap: gravity does act on the bar, but at horizontal equilibrium, the gravitational moment is already balanced by static spring preload, so gravity contributes no additional linearized restoring stiffness for small oscillations. Strategy tip: For any rotating rigid body, always convert everything into the rotational domain — find $k_{\text{eff}} = kd^2$ (where $d$ is the moment arm) and use the correct $I$ about the pivot. The $L^2$ terms will cancel cleanly, but only if you set the problem up correctly from the start.
3. Which of the following best evaluates the student's argument regarding the effect of gravity on the natural frequency of the vertically suspended mass–spring system?
   Answer: The student is wrong. When the equation of motion is written in terms of displacement measured from the static equilibrium position, the weight $mg$ and the static spring force cancel exactly, leaving $m\ddot{x} + kx = 0$ and thus $\omega_n = \sqrt{k/m}$, unchanged by gravity regardless of the magnitude of the static deflection. Whenever you see a question about vertical spring-mass systems, your first instinct should be to write out the full equation of motion carefully — don't reason about gravity's effect qualitatively before doing the math. Here's why B is correct. Let $x$ be displacement measured from the static equilibrium position, where the spring is already stretched by $\delta_{st} = mg/k$. When the mass moves an additional distance $x$ downward, the net spring force upward is $k(\delta_{st} + x)$, and gravity pulls down with $mg$. Newton's second law gives: $m\ddot{x} = mg - k(\delta_{st} + x) = mg - k\delta_{st} - kx$ Since $k\delta_{st} = mg$, these terms cancel exactly, leaving $m\ddot{x} + kx = 0$, so $\omega_n = \sqrt{k/m}$ — gravity drops out entirely. A is wrong because it invents a correction term $\sqrt{(k - mg/\delta_{st})/m}$. Notice that $mg/\delta_{st} = k$ always, so this formula would give zero — a nonsensical result born from faulty reasoning about gravity "offsetting" the spring force during oscillation. C reaches the right numerical answer but for the wrong reason. Saying "gravity doesn't change $k$ or $m$" skips the real issue: gravity does appear in the equation of motion — it just cancels. The cancellation is the insight, not a shortcut around it. D is wrong in both claim and conclusion. The downward shift of equilibrium does not increase effective stiffness; stiffness is a property of the spring alone. Study tip: Always write the equation of motion in terms of displacement measured from static equilibrium — this coordinate choice is what makes gravity vanish, and it's a standard technique you should apply automatically on exam questions involving vertical oscillation.
4. What is the natural frequency of oscillation of the rolling disk–spring system?
   Answer: $\omega_n = \sqrt{\dfrac{2k}{3m}}$, because the rolling constraint introduces an effective inertia equal to $\frac{3}{2}m$ (the translational mass plus the rotational contribution $\frac{1}{2}m$ from the disk's moment of inertia $\frac{1}{2}mR^2$), leaving the stiffness $k$ unchanged. When a disk rolls without slipping, you must account for both translational and rotational inertia — this is the central concept being tested. The system has one degree of freedom (the center displacement $x$), but the disk's rotation is kinematically linked to its translation via the no-slip condition $\dot{\theta} = \dot{x}/R$. To find the equation of motion, write the energy or apply Newton's laws with torques. The kinetic energy is $T = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\dot{\theta}^2$, where $I = \frac{1}{2}mR^2$ for a disk. Substituting the rolling constraint gives $T = \frac{1}{2}m\dot{x}^2 + \frac{1}{4}m\dot{x}^2 = \frac{3}{4}m\dot{x}^2$. The potential energy is simply $V = \frac{1}{2}kx^2$. Applying Lagrange's equation yields $\frac{3}{2}m\ddot{x} + kx = 0$, so the natural frequency is $\omega_n = \sqrt{\frac{2k}{3m}}$, confirming C. A is wrong because ignoring rotational inertia treats the disk as a sliding block — the rolling constraint is physically significant and cannot be dropped. B correctly senses that rolling adds inertia, but doubles the full mass rather than adding the correct rotational contribution $\frac{1}{2}m$, giving an effective inertia of $2m$ instead of $\frac{3}{2}m$. D is wrong because the no-slip condition constrains kinematics, not stiffness — the spring still stretches by $x$, so $k$ remains unchanged. Study tip: For any rolling-body oscillation problem, your first step should always be writing the full kinetic energy including the $\frac{1}{2}I\omega^2$ term, then substituting the rolling constraint. This systematically gives the correct effective inertia every time.
5. What is the ratio of the natural frequency after the base modification to the natural frequency before, assuming the floor mass $m$ is unchanged and both columns act in parallel for lateral motion?
   Answer: $\dfrac{\omega_{n,\text{after}}}{\omega_{n,\text{before}}} = \dfrac{1}{2}$, because the two columns in parallel give a total stiffness reduced by a factor of 4 (from $2k_c$ to $2k_c/4 = k_c/2$), and natural frequency scales as the square root of stiffness, so the ratio is $\sqrt{1/4} = 1/2$. When a floor mass is supported by lateral springs (columns) acting in parallel, the total lateral stiffness is simply the sum of the individual column stiffnesses. Natural frequency follows $\omega_n = \sqrt{k_{\text{total}}/m}$, so it scales as the square root of total stiffness. Keep both of these relationships in mind as you work through the modification. Before the modification, both columns are fixed-fixed, each with stiffness $k_c$, giving $k_{\text{before}} = k_c + k_c = 2k_c$. After pinning both bases, each column's stiffness drops to $k_c/4$, so $k_{\text{after}} = k_c/4 + k_c/4 = k_c/2$. The stiffness ratio is therefore $k_{\text{after}}/k_{\text{before}} = (k_c/2)/(2k_c) = 1/4$. Taking the square root gives the frequency ratio: $\omega_{n,\text{after}}/\omega_{n,\text{before}} = \sqrt{1/4} = 1/2$. Answer D is correct. Answer A incorrectly pins only one column while leaving the other fixed, contradicting the problem statement that both bases are modified. Answer B makes the classic error of confusing stiffness ratios with frequency ratios — frequency scales as the square root of stiffness, not linearly, so a fourfold stiffness reduction yields only a twofold frequency reduction. Answer C invents a fictitious extra $\sqrt{2}$ penalty for parallel springs; parallel combination is simply additive, with no additional correction factor. A reliable study tip: whenever a spring-mass system is modified, compute $k_{\text{total}}$ before and after explicitly, then take the square root of their ratio to find the frequency ratio. Never skip the square root step — it's the most common arithmetic trap on dynamics questions.
6. Which expression correctly gives the natural frequency of small oscillations of this pendulum–spring system?
   Answer: $\omega_n = \sqrt{\dfrac{ka^2 + mgL/2}{mL^2/3}}$, because both the spring restoring moment ($ka^2\theta$) and the gravitational restoring moment about the pin ($mg(L/2)\theta$ for small $\theta$) act to return the rod to vertical, and the rotational inertia of the rod about the pin is $mL^2/3$. When a rod is pinned at one end and free to rotate, you should immediately reach for the equation of rotational motion: sum of moments about the pin equals $I_{\text{pin}} \cdot \ddot{\theta}$. The key is correctly identifying every moment acting on the system and the right rotational inertia. For a slender rod pinned at its upper end, the rotational inertia about the pin is $I_{\text{pin}} = mL^2/3$ (via the parallel-axis theorem: $mL^2/12 + m(L/2)^2$). When displaced by small angle $\theta$, the spring at distance $a\$ stretches by$a\theta$, producing a restoring moment$$ka^2$\theta$. Critically, gravity also contributes: the rod's weight$mg$acts at the center of mass$(L/2)$from the pin, and for a hanging rod displaced from vertical, this force creates a **restoring** moment of$mg(L/2)\theta$pulling it back toward equilibrium. The equation of motion becomes$$mL^2$/3 \cdot \ddot{\theta} = -($ka^2$ + mgL/2)\theta$, yielding$\omega_n = \sqrt{($ka^2$ + mgL/2)/($mL^2$/3)}$. That's answer **D**. **A** uses$$mL^2$/12$$, the inertia about the center of mass — this is only correct if the rod were rotating about its midpoint, not the pin. B wrongly claims gravity produces no restoring moment; while gravity is vertical and the rod is vertical at equilibrium, a displaced rod absolutely develops a gravitational restoring torque. C subtracts the gravity term, confusing this hanging pendulum (gravity restores) with an inverted pendulum (gravity destabilizes). As a study habit: always sketch the displaced configuration, identify every force, and determine whether each moment aids or opposes return to equilibrium before writing your equation of motion.
7. By what factor does the natural frequency change as a result of both modifications applied together?
   Answer: The natural frequency increases by a factor of $\sqrt{2}$, because the stiffness quadruples (factor of 4) and the mass doubles (factor of 2), so $\omega_n' / \omega_n = \sqrt{4/2} = \sqrt{2}$. Whenever you see a question about how modifications affect natural frequency, your anchor formula is $\omega_n = \sqrt{k/m}$. Both stiffness and mass appear under the same radical, so you must account for both changes simultaneously — never in isolation. Starting from the original system, the new natural frequency after both modifications is: $\omega_n' = \sqrt{\frac{4k}{2m}} = \sqrt{\frac{4}{2} \cdot \frac{k}{m}} = \sqrt{2}\,\omega_n$ So the natural frequency increases by a factor of $\sqrt{2}$, confirming that A is correct. Here's where the distractors go wrong. B claims that mass plays no role in the dynamic equation — this is flatly false. Mass absolutely appears in the equation of motion ($m\ddot{x} + kx = 0$), and increasing it lowers natural frequency. Ignoring the mass doubling gives an inflated result of 2. C compounds the error by treating the two effects as independent multipliers that add rather than combine under a single radical. The stiffness quadrupling contributes a factor of $\sqrt{4} = 2$, and the mass doubling contributes a factor of $1/\sqrt{2}$, giving $2 \times (1/\sqrt{2}) = \sqrt{2}$ — not $2\sqrt{2}$. D claims the effects cancel, which would require the stiffness and mass to change by the same factor; here stiffness grows by 4 and mass by 2, so there is a net change. A useful habit: always write the ratio $\omega_n'/\omega_n = \sqrt{(k'/k)/(m'/m)}$ and plug in the multipliers before doing any arithmetic. This keeps both changes visible and prevents the partial-substitution errors seen in B, C, and D.
8. The mass is displaced a distance $x$ to the right and released. Which of the following correctly identifies the equation of motion and the resulting natural frequency?
   Answer: Equation of motion: $m\ddot{x} + 3kx = 0$; natural frequency $\omega_n = \sqrt{3k/m}$, because when the mass moves right by $x$, the left spring pulls left with force $kx$ and the right spring pushes left with force $2kx$, giving a total restoring force of $3kx$ and an effective stiffness of $3k$. When a mass sits between two walls connected by springs, your first instinct should be to draw a free-body diagram and carefully track the direction of every spring force after the mass is displaced. When the mass moves right by $x$, consider what each spring does. The left spring (stiffness $k$) is stretched, so it pulls the mass back to the left with force $kx$. The right spring (stiffness $2k$) is compressed, so it pushes the mass back to the left with force $2kx$. Both forces act in the same direction — opposing the displacement. Summing them gives a total restoring force of $3kx$, so Newton's second law yields $m\ddot{x} = -3kx$, or $m\ddot{x} + 3kx = 0$, with natural frequency $\omega_n = \sqrt{3k/m}$. This is why B is correct — the two springs act in parallel from the mass's perspective, and their stiffnesses add directly. A is wrong because it assumes the right spring goes slack, as if it were a rope. A compressed spring absolutely exerts a force — it doesn't go slack, it pushes back. C applies the series stiffness formula, which is only valid when two springs share the same tension (e.g., chained end-to-end with no intermediate mass). Here the mass is between the springs, so they act in parallel, not series. D invents a fictional "dominant spring" principle. There is no such rule in dynamics — both springs do work throughout the motion, and their contributions don't cancel. As a study habit: whenever a mass is anchored between two supports, expect parallel springs and add the stiffnesses.
9. Two identical springs, each with stiffness $k$, are used to support a mass $m$. In Configuration 1, both springs connect the mass to a rigid wall in parallel (the mass moves and both springs stretch/compress simultaneously). In Configuration 2, the two springs are arranged in series between the mass and the same rigid wall. A student claims that the ratio of natural frequencies satisfies $\omega_{n,1}/\omega_{n,2} = 2$. Is the student correct, and what is the actual ratio?
   Answer: The student is correct. The parallel arrangement doubles the stiffness relative to the series arrangement, so the frequency ratio equals $\sqrt{2k / (k/2)} = 2$, confirming the claim exactly. When analyzing spring-mass systems, your first step should always be to find the equivalent stiffness of the spring arrangement, since natural frequency depends on it via $\omega_n = \sqrt{k_{eq}/m}$. For parallel springs, both springs share the same displacement, so their forces add: $k_{eq,parallel} = k + k = 2k$. For series springs, both springs carry the same force but their displacements add, giving $k_{eq,series} = \frac{k \cdot k}{k + k} = \frac{k}{2}$. The stiffness ratio is therefore $\frac{k_{eq,1}}{k_{eq,2}} = \frac{2k}{k/2} = 4$, and since frequency scales as the square root of stiffness, the frequency ratio is $\frac{\omega_{n,1}}{\omega_{n,2}} = \sqrt{4} = 2$. The student's numerical answer of 2 is exactly right, and their reasoning — that parallel stiffness is $2k$ and series is $k/2$, yielding a ratio of 4 under the square root — is also correct. Answer A is right. Answer B claims the reasoning is flawed, but the student's logic is actually sound — both the stiffness values and the square-root step are correctly applied, so B's critique is invalid. Answer C incorrectly states that series springs with equal stiffness yield $k_{eq} = k$; this confuses series springs with a single spring. The correct formula gives $k/2$, not $k$. Answer D invents an error about the ratio being "inverted" — the question clearly defines the ratio as $\omega_{n,1}/\omega_{n,2}$, which is 2, not $1/2$. Study tip: Always derive $k_{eq}$ before computing frequency — and remember that series springs are weaker ($k_{eq} < k$), while parallel springs are stiffer ($k_{eq} > k$).
10. After removing one spring, what is the new natural frequency of the vertical vibration, assuming the mass distribution and boundary conditions are otherwise unchanged?
    Answer: $f_n' = 5\sqrt{\dfrac{3}{4}} \approx 4.33\text{ Hz}$, because removing one of four parallel springs reduces the total stiffness from $4k$ to $3k$, and natural frequency scales as the square root of stiffness. Whenever you see a vibration problem involving springs, anchor yourself to the fundamental formula: $f_n = \frac{1}{2\pi}\sqrt{\frac{k_{total}}{M}}$. This tells you natural frequency scales as the square root of total stiffness — a relationship that governs every distractor in this question. When four identical springs of stiffness $k$ are arranged in parallel (each spring shares the load side-by-side), their stiffnesses add directly: $k_{total} = 4k$. Removing one spring reduces total stiffness to $3k$. The ratio of new to old frequency is therefore: $\frac{f_n'}{f_n} = \sqrt{\frac{3k}{4k}} = \sqrt{\frac{3}{4}}$ So $f_n' = 5\sqrt{\frac{3}{4}} \approx 4.33 \text{ Hz}$, confirming A is correct. B makes the classic mistake of treating frequency as linearly proportional to stiffness — scaling by $3/4$ directly instead of taking the square root. The square root relationship is non-negotiable here. C inverts the ratio, claiming frequency increases after removing a spring. This contradicts physics: fewer parallel springs means lower total stiffness, which always lowers natural frequency. The reasoning about "larger deflections increasing stiffness" is fabricated — individual spring constants don't change based on load sharing. D compounds two errors: it inverts the ratio and treats the relationship as linear, producing a frequency increase proportional to $4/3$. No physical principle supports this. Your go-to study tip: always write $f_n \propto \sqrt{k}$ before solving any spring-mass frequency problem. This single relationship eliminates B and D (wrong proportionality) and forces you to check whether stiffness increases or decreases, catching C as well.

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