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For the damped oscillator shown in Figure the mass mof 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,
$(b)$ time taken for its amplitude of vibrations to drop to half of Its inittal value, and
$(c)$ the time taken for its mechanical energy to drop to half its initial value.
Solution
$(a)$ We see that $k m=90 \times 0.2=18\, kg \,N$
$m ^{-1}= kg ^{2}\, s ^{-2} ;$
therefore $\sqrt{k m}=4.243\, kg \,s ^{-1},$ and $b=0.04\, kg\, s ^{-1} .$
Therefore, $b$ is much less than $\sqrt{k m} .$ Hence, the time pertod $T$
$T=2 \pi \sqrt{\frac{m}{k}}$
$=2 \pi \sqrt{\frac{0.2\, kg }{90 \,N m ^{-1}}}$
$=0.3 \,s$
$(b)$ the time, $T_{1 / 2}$, for the amplitude to drop to half of 1ts inttlal value is given by,
$T_{1 / 2}=\frac{\ln (1 / 2)}{b / 2 m}$
$=\frac{0.693}{40} \times 2 \times 200\, s$
$=6.93 \,s$
$(c)$ For calculating the time, $t_{1 / 2},$ for its mechanical energy to drop to half its initial value
$E\left(t_{1 / 2}\right) / E(0)=\exp \left(-b t_{1 / 2} / m\right)$
$\frac 12 =\exp^{-\left(\frac{b t_{\frac 12}}{ m}\right)}$
$\ln (\frac 12)=-\left(\frac{b t_{\frac 12}}{ m}\right)$
$t_{\frac 12}=\frac{0.693}{40 \,g \,s ^{-1}} \times 200\, g$
$=3.46 \,s$
This is just half of the decay period for amplitude. This is not surprising, because. energy depends on the square of the amplitude. Notice that there is a factor of $2$ in the exponents of the two exponentials.
