$A$ damped harmonic oscillator has a frequency of $5$ oscillations per second. The amplitude drops to half its value for every $10$ oscillations. The time it will take to drop to $\frac{1}{1000}$ of the original amplitude is close to .... $s$

The amplitude of a simple pendulum,oscillating in air with a small spherical bob,decreases from $10 \ cm$ to $8 \ cm$ in $40 \ s$. Assuming that Stokes' law is valid,and the ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$. The time in which the amplitude of this pendulum will reduce from $10 \ cm$ to $5 \ cm$ in carbon dioxide will be close to ..... $s$ $(\ln 5 = 1.601, \ln 2 = 0.693)$

What are damped oscillations? Discuss them using the illustration of a spring.

$A$ spring-mass system executes damped harmonic oscillations given by the equation $y = A e^{-\frac{bt}{2m}} \sin(\omega' t + \phi)$,where the symbols have their usual meanings. If a $2 \ kg$ mass $(m)$ is attached to a spring of force constant $(K) = 1250 \ N/m$,the period of the oscillation is $(\pi / 12) \ s$. The damping constant $b$ has the value ..... $kg/s$.

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?

Four pendulums $A, B, C$ and $D$ are hung from the same elastic support as shown in the figure. $A$ and $C$ are of the same length,while $B$ is shorter than $A$ and $C$,and $D$ is longer than $A$. If $A$ is given a displacement,then at steady state: