Two damped spring-mass oscillating systems have identical spring constants and decay times. However,system $A$'s mass $m_A$ is twice system $B$'s mass $m_B$. How do their damping constants,$b$,compare?

If a simple pendulum has significant amplitude (up to a factor of $1/e$ of original) only in the period between $t = 0 \ s$ to $t = \tau \ s$,then $\tau$ may be called the average life of the pendulum. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $b$ as the constant of proportionality,the average life time of the pendulum is (assuming damping is small) in seconds:

$A$ body of mass $0.3 \ kg$ hangs by a spring with a force constant of $50 \ N/m$. The amplitude of oscillations is damped and reaches $1/e$ of its original value in about $100$ oscillations. If $\omega$ and $\omega^{\prime}$ are the angular frequencies of undamped and damped oscillations respectively, then the percentage value of $\left(\frac{\omega-\omega^{\prime}}{\omega}\right) \times 100$ is:

The displacement of a damped harmonic oscillator is given by $x(t) = e^{-0.1t} \cos(10\pi t + \varphi)$. The time taken for its amplitude of vibration to drop to half of its initial value is close to .... $s$

The angular frequency of the damped oscillator is given by $\omega = \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}}$,where $k$ is the spring constant,$m$ is the mass of the oscillator,and $r$ is the damping constant. If the ratio $\frac{r^2}{mk}$ is $8\%$,the change in time period compared to the undamped oscillator is approximately as follows:

$A$ simple pendulum oscillates in air with time period $T$ and amplitude $A$. As the time passes,