Statics and Dynamics Quiz: Mass Spring Systems

What this quiz covers

This quiz focuses on Mass Spring Systems, 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. $f_n' = 5\sqrt{\dfrac{3}{4}} \approx 4.33\text{ Hz}$, 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. (correct answer)
2. $f_n' = 5 \times \dfrac{3}{4} = 3.75\text{ Hz}$, because the stiffness decreases by a factor of $3/4$ when one spring is removed from the parallel arrangement, and natural frequency scales linearly with stiffness.
3. $f_n' = 5\sqrt{\dfrac{4}{3}} \approx 5.77\text{ Hz}$, because removing one spring forces the remaining three springs to carry a larger share of the load, effectively increasing their individual deflections and thus raising the system stiffness per unit mass.
4. $f_n' = 5 \times \dfrac{4}{3} \approx 6.67\text{ Hz}$, because the natural frequency is inversely proportional to the number of springs, so reducing the spring count from four to three increases the frequency by the ratio $4/3$.

Question 2

1. $\omega_n = \sqrt{\dfrac{ka^2 + mgL/2}{mL^2/12}}$, because the rotational inertia of a slender rod about its center of mass is $mL^2/12$, and this is the appropriate inertia to use when the system's center of mass oscillates about the pin.
2. $\omega_n = \sqrt{\dfrac{ka^2}{mL^2/3}}$, because gravity acts through the center of mass along the rod axis and produces no moment about the pin for vertical equilibrium, so only the spring provides a restoring moment for small oscillations.
3. $\omega_n = \sqrt{\dfrac{ka^2 - mgL/2}{mL^2/3}}$, because for a hanging pendulum displaced from vertical, gravity acts as a destabilizing (anti-restoring) moment that reduces the effective stiffness, so the gravitational term must be subtracted from the spring stiffness contribution.
4. $\omega_n = \sqrt{\dfrac{ka^2 + mgL/2}{mL^2/3}}$, 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 $mL^2/3$. (correct answer)

Question 3

1. The colleague is incorrect. Increasing mass by factor $n$ gives frequency $\omega_n/\sqrt{n}$, while reducing stiffness to $k/n^2$ gives frequency $\omega_n/n$. Because $\omega_n/n < \omega_n/\sqrt{n}$ for $n > 1$, the spring strategy reaches a lower frequency than intended, but this makes it a conservative (safer) design choice, so the colleague's suggestion is acceptable in practice.
2. The colleague is correct. Increasing mass by factor $n$ gives frequency $\omega_n/\sqrt{n}$, and reducing stiffness to $k/n^2$ gives frequency $\omega_n/\sqrt{n^2} = \omega_n/n$. These are equal because $\sqrt{n^2} = n$, confirming the two strategies produce the same natural frequency for any value of $n$.
3. 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 $k/n^2$. (correct answer)
4. The colleague is correct. Both strategies yield the same natural frequency because multiplying $m$ by $n$ is algebraically equivalent to dividing $k$ by $n^2$, since the relationship $\omega_n \propto \sqrt{k/m}$ means that any combination of $k$ and $m$ satisfying $k/m = \text{constant}$ produces the same frequency, and $k/(nm) = (k/n^2)/m$ satisfies this condition.

Question 4

1. $\omega_n = \sqrt{\dfrac{k}{m}}$, because the spring force and the bar mass are the only dynamic quantities, and the rotational inertia of the bar does not alter the equilibrium stiffness seen by the system.
2. $\omega_n = \sqrt{\dfrac{3k}{m}}$, 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 $\frac{kL^2}{mL^2/3}$. (correct answer)
3. $\omega_n = \sqrt{\dfrac{k}{3m}}$, because the effective mass seen by the spring at the tip of the bar is one-third of the total bar mass due to the distributed inertia, so the system reduces to a spring–mass pair with $m_{\text{eff}} = 3m$.
4. $\omega_n = \sqrt{\dfrac{4k}{3m}}$, because the gravitational restoring moment must be added to the spring moment when linearizing about the horizontal equilibrium, increasing the effective stiffness beyond $kL^2$.

Question 5

1. Equation of motion: $m\ddot{x} + kx = 0$; natural frequency $\omega_n = \sqrt{k/m}$, because only the left spring (stiffness $k$) provides a restoring force when the mass moves right; the right spring (stiffness $2k$) goes slack and contributes nothing.
2. 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 $3k$. (correct answer)
3. Equation of motion: $m\ddot{x} + kx = 0$; natural frequency $\omega_n = \sqrt{k/m}$, because the two springs are in series between the two walls, and the series combination of $k$ and $2k$ yields an effective stiffness of $2k/3$, which then simplifies to $k$ for this geometry.
4. Equation of motion: $m\ddot{x} + 2kx = 0$; natural frequency $\omega_n = \sqrt{2k/m}$, because the dominant spring is the stiffer one ($2k$), and by the principle of virtual work only the stiffer spring does net work during displacement while the weaker spring's contribution averages to zero over a full cycle.

Question 6

1. The student is partially correct: gravity introduces an anti-restoring effect that reduces the natural frequency to $\omega_n = \sqrt{(k - mg/\delta_{st})/m}$, where $\delta_{st}$ is the static deflection, because the weight partially offsets the spring force during oscillation. This correction becomes significant when $mg$ is comparable to $k\delta_{st}$.
2. 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. (correct answer)
3. The student is wrong, but for a trivial reason: since natural frequency depends only on the ratio $k/m$, and gravity alters neither the spring constant $k$ nor the mass $m$, the frequency $\sqrt{k/m}$ holds automatically. No examination of the equation of motion is necessary to reach this conclusion.
4. The student is correct: gravity shifts the equilibrium position downward by $\delta_{st} = mg/k$, and this downward shift increases the effective spring stiffness experienced during oscillation, raising the natural frequency above $\sqrt{k/m}$ for larger values of static deflection.

Question 7

1. 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. (correct answer)
2. The student is incorrect. The parallel stiffness is $2k$ and the series stiffness is $k/2$, giving a stiffness ratio of 4 and a frequency ratio of $\sqrt{4} = 2$, so the student's numerical answer is right but the reasoning about doubling stiffness is wrong.
3. The student is incorrect. The parallel stiffness is $2k$ and the series stiffness is $2k$, because series springs with equal stiffness yield an equivalent stiffness equal to the individual stiffness. The frequency ratio is therefore 1, not 2.
4. The student is incorrect. The parallel stiffness is $2k$ and the series stiffness is $k/2$, giving a frequency ratio of $\sqrt{2k/(k/2)} = 2$, which equals the student's answer numerically. However, the ratio should be computed as $\omega_{n,2}/\omega_{n,1}$, which is $1/2$, so the student has the ratio inverted.

Question 8

1. 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 $\omega_n' / \omega_n = \sqrt{4/2} = \sqrt{2}$. (correct answer)
2. The natural frequency increases by a factor of $2$, because quadrupling the stiffness raises the frequency by a factor of $\sqrt{4} = 2$, and the mass doubling does not affect the result since mass appears only in the static equilibrium, not the dynamic equation.
3. The natural frequency increases by a factor of $2\sqrt{2}$, because the stiffness change multiplies the frequency by $\sqrt{4} = 2$ and the mass change independently multiplies it by $\sqrt{2}$, so the combined effect is $2 \times \sqrt{2}$.
4. The natural frequency remains unchanged, because quadrupling the stiffness and doubling the mass both alter $\omega_n$ by the same proportional amount in opposite directions, and the net effect on $\omega_n = \sqrt{k/m}$ is a cancellation.

Question 9

1. $\dfrac{\omega_{n,\text{after}}}{\omega_{n,\text{before}}} = \dfrac{1}{\sqrt{2}}$, because the stiffness reduction applies only to one of the two columns (the other remains fixed at both ends), so the total stiffness decreases from $2k_c$ to $k_c + k_c/4 = 5k_c/4$, giving a frequency ratio of $\sqrt{(5/4)/(2)} = 1/\sqrt{2}$ approximately.
2. $\dfrac{\omega_{n,\text{after}}}{\omega_{n,\text{before}}} = \dfrac{1}{4}$, because each column's stiffness is reduced by a factor of 4, and since frequency is directly proportional to stiffness in a spring–mass system, the natural frequency also decreases by a factor of 4.
3. $\dfrac{\omega_{n,\text{after}}}{\omega_{n,\text{before}}} = \dfrac{1}{2\sqrt{2}}$, because the parallel combination of two springs reduces the effective stiffness by an additional factor of $\sqrt{2}$ compared to a single spring, which must be accounted for in the frequency ratio calculation.
4. $\dfrac{\omega_{n,\text{after}}}{\omega_{n,\text{before}}} = \dfrac{1}{2}$, 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 $\sqrt{1/4} = 1/2$. (correct answer)

Question 10

1. $\omega_n = \sqrt{\dfrac{k}{m}}$, because the spring force acts at the center of mass, and for a rolling constraint the translational equation of motion $m\ddot{x} + kx = 0$ fully describes the system dynamics without any rotational correction.
2. $\omega_n = \sqrt{\dfrac{k}{2m}}$, because rolling without slipping means the contact point is instantaneously at rest, effectively doubling the mass that the spring must accelerate: the disk's full mass $m$ plus an equal rotational equivalent mass $m$.
3. $\omega_n = \sqrt{\dfrac{2k}{3m}}$, 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. (correct answer)
4. $\omega_n = \sqrt{\dfrac{2k}{m}}$, because the no-slip rolling constraint forces the contact point to remain stationary, which acts as an additional fixed point that doubles the effective spring stiffness seen by the disk's center of mass while leaving the inertia unchanged.