A linear single-degree-of-freedom oscillator has mass m=1 kg, viscous damping c=6…

A linear single-degree-of-freedom oscillator has mass $m=1\ \text{kg}$, viscous damping $c=6\ \text{N}\cdot\text{s/m}$, and stiffness $k=25\ \text{N/m}$. There is no external force. The free motion satisfies \[ \ddot x+6\dot x+25x=0, \] with the rest-release data \[ x(0)=0.1\ \text{m},\qquad \dot x(0)=0. \] 1. Compute the undamped natural frequency $\omega_n$, the damping ratio $\zeta$, and the damped frequency $\omega_d$. Confirm that the motion is underdamped. 2. Write the underdamped free solution \[ x(t)=e^{-\zeta\omega_n t}\bigl(A\cos\omega_d t+B\sin\omega_d t\bigr) \] and determine $A$ and $B$ from the initial data. (A vanishing initial velocity does not force $B=0$.) 3. Compute the damped period $T_d=2\pi/\omega_d$ and the logarithmic decrement \[ \delta=\ln\frac{x(t)}{x(t+T_d)}=\zeta\omega_n T_d \] evaluated on successive same-phase peaks. 4. Report the same-phase peak ratio $x(t)/x(t+T_d)$ as an exact exponential. 5. Audit both sentences: (i) “because the mass is released from rest, the solution is a pure damped cosine, $B=0$”; (ii) “the logarithmic decrement is the decay constant $\zeta\omega_n$ itself.” The oscillator is free. Do not introduce a harmonic forcing function, a particular solution, or a two-mass modal analysis.

$\omega_n=\sqrt{25}=5$ and $\zeta=6/(2\cdot 5)=3/5<1$, so $\omega_d=5\sqrt{1-9/25}=4$.