For the damped oscillator shown in the figure,the mass $m$ of the block is $200 \; g$,$k = 90 \; N m^{-1}$,and the damping constant $b$ is $40 \; g s^{-1}$. Calculate:
$(a)$ the period of oscillation,
$(b)$ the time taken for its amplitude of vibrations to drop to half of its initial value,and
$(c)$ the time taken for its mechanical energy to drop to half its initial value.

(N/A) Given: $m = 200 \; g = 0.2 \; kg$,$k = 90 \; N m^{-1}$,$b = 40 \; g s^{-1} = 0.04 \; kg s^{-1}$.
We check the condition for light damping: $\sqrt{km} = \sqrt{90 \times 0.2} = \sqrt{18} \approx 4.243 \; kg s^{-1}$.
Since $b \ll \sqrt{km}$,the period of oscillation $T$ is given by:
$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.2}{90}} = 2\pi \sqrt{\frac{1}{450}} = \frac{2\pi}{21.21} \approx 0.296 \; s \approx 0.3 \; s$.
$(b)$ The amplitude $A(t)$ decays as $A(t) = A_0 e^{-bt/2m}$. For $A(t) = A_0/2$:
$\frac{1}{2} = e^{-bt/2m} \implies \ln(2) = \frac{bt}{2m} \implies t = \frac{2m \ln(2)}{b}$.
$t = \frac{2 \times 0.2 \times 0.693}{0.04} = \frac{0.2772}{0.04} = 6.93 \; s$.
$(c)$ The mechanical energy $E(t)$ decays as $E(t) = E_0 e^{-bt/m}$. For $E(t) = E_0/2$:
$\frac{1}{2} = e^{-bt/m} \implies \ln(2) = \frac{bt}{m} \implies t = \frac{m \ln(2)}{b}$.
$t = \frac{0.2 \times 0.693}{0.04} = \frac{0.1386}{0.04} = 3.465 \; s \approx 3.46 \; s$.