Because $x$ is measured from static equilibrium, where $k\delta_{st}=mg$. Those two terms cancel. Gravity only shifts the equilibrium position.

Parallel: $k_{eq}=2k$. Series: $k_{eq}=k/2$. Since $\omega_n\propto\sqrt{k_{eq}}$, the ratio is $\omega_{par}/\omega_{ser}=\sqrt{2k\,/\,(k/2)}=\mathbf{2}$.

$\sum M_O=I_O\ddot\theta$ with $I_O=\tfrac13 mL^2$ and gravity acting at $L/2$: $\tfrac13 mL^2\ddot\theta=-mg\tfrac L2\sin\theta$. Linearizing gives $\ddot\theta+\frac{3g}{2L}\theta=0$, so $\omega_n=\sqrt{3g/2L}$. That is $\sqrt{1.5}\approx1.22$ times faster than a simple pendulum, because more of the rod's mass sits close to the pivot.

$k\delta_{st}=mg$ gives $\omega_n=\sqrt{k/m}=\sqrt{g/\delta_{st}}=\sqrt{9.81/0.02}\approx22$ rad/s, so $f_n\approx\mathbf{3.5}$ Hz. You don't need $k$ or $m$ separately, only the static deflection.

$\sin\theta\approx\theta$ has error $\approx\theta^3/6$. The real pendulum's restoring moment is weaker than the linear one at large angles, so it swings more slowly and its period grows with amplitude. A linear system's period never depends on amplitude.

$f_n\propto\sqrt{k/m}$, so quadrupling $m$ halves $f_n$: 1 Hz.

$\omega_n=2\pi(2)=12.6$ rad/s. $v_{max}=\omega_nX=\mathbf{0.63}$ m/s and $a_{max}=\omega_n^2X=\mathbf{7.9}$ m/s². Each derivative multiplies the amplitude by $\omega_n$. (Equivalently, energy: $\tfrac12 kX^2=\tfrac12 mv_{max}^2$.)

$\delta=\ln(10/8)=0.223$, so $\zeta=\delta/\sqrt{4\pi^2+\delta^2}\approx\mathbf{0.036}$.

$\omega_d/\omega_n=\sqrt{1-\zeta^2}$: 0.995 for $\zeta=0.1$ (a 0.5% change) and 0.866 for $\zeta=0.5$. Light damping barely affects the frequency, but it strongly controls how fast the motion decays.

$\zeta=1$. Overdamped systems are dominated by the slow root $s_1=-\omega_n(\zeta-\sqrt{\zeta^2-1})$, which gets closer to zero as $\zeta$ grows. Check with the comparison plot in Section 3.

A door should shut as fast as possible without slamming or bouncing, which is exactly what critical damping does. A car suspension must also isolate passengers from road bumps; lower damping transmits less of the high-frequency road input (see Section 5), at the cost of a little overshoot after each bump.

$\delta=\ln(4/5)=-0.223$, which gives $\zeta\approx\mathbf{-0.036}$. Negative damping means something is feeding energy into the system on every cycle, as in aircraft wing flutter, brake squeal, or a bowed violin string. Try $\zeta<0$ in Section 3.

The displacement lags the force by 90°. The force is then in phase with the velocity, so it is balanced entirely by the damping force; only $c$ limits the amplitude ($X=\delta_{st}/2\zeta$).

$M\approx1/2\zeta=\mathbf{10}$. A force that would deflect the structure 1 mm if applied slowly produces about 10 mm of vibration at resonance.

Beats: the free and forced responses drift in and out of step. The envelope repeats at the difference frequency $|f-f_n|=0.5$ Hz, so the amplitude swells every 2 s. Try the Beats preset in Section 4.

The transient decays as $e^{-\zeta\omega_n t}$ with time constant $1/\zeta\omega_n=2$ s. After about four time constants it is down to 2%, so about 8 s. Turn on Transient in Section 4 to watch it fade.

In the mass-controlled region $M\approx1/r^2$, so doubling $\omega$ cuts the amplitude by about a factor of 4. The mass is too heavy to follow a faster force.

By superposition: because the equation is linear, find the steady-state response to each term separately (each with its own $M$ and $\phi$) and add them. This is what makes Fourier analysis of vibration possible, and it fails for nonlinear systems like the large-angle pendulum.

Need $r>\sqrt2$, so $f_n<30/\sqrt2\approx21$ Hz; in practice aim for $r\approx3$ or more ($f_n\approx10$ Hz or lower). Above $r=\sqrt2$, more damping increases transmissibility, so use only enough damping to limit motion when passing through resonance at startup.

The base excitation frequency is $f=v/\lambda$. Resonance when $f=f_n$: $v=f_n\lambda=1.5\times5=\mathbf{7.5}$ m/s (27 km/h, 17 mph). Driving much faster puts the car in the isolation region, which is why washboard roads can feel smoother at higher speed.

A soft spring gives a low $\omega_n$, so earthquake frequencies have $r\gg1$. Then $TR\to0$: the mass stays almost still in space while the case moves with the ground, so the relative motion between case and mass equals the ground displacement. An accelerometer uses the opposite limit (very stiff, $r\ll1$), where the relative motion is proportional to the base acceleration.

$K$ has the opposite sign to $M$, so $\omega_n=\sqrt{K/M}$ is imaginary: the solution grows exponentially without oscillating. A passive mass–spring–dashpot system cannot do that, so there is a derivation error, most likely a spring force drawn in the wrong direction on an FBD or a sign slip in the kinematics. They should recheck the FBD (all coordinates positive, then stretched or compressed?) instead of just flipping the sign.

Well below resonance the amplitude is about $me\omega^2/k$, so doubling $\omega$ multiplies it by about 4. Well above resonance $X\to me/(M+m)$ regardless of speed, so the amplitude hardly changes. The growing force is exactly cancelled by the mass becoming harder to move.