1. Tardigrade
  2. Question
  3. Physics
  4. A block of mass 200 g executing SHM under the influence of a spring of spring constant k = 90 N m-1 and a damping constant b = 40 g s-1. The time elapsed for its amplitude to drop to half of its initial value is (Given, In (l/2) = -0.693), the time elapsed for its mechanical energy to drop half of its initial value is

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Q. A block of mass $200\, g$ executing $SHM$ under the influence of a spring of spring constant $k = 90 \,N\, m^{-1}$ and a damping constant $b = 40 \,g\, s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is (Given, In (l/2) = -0.693), the time elapsed for its mechanical energy to drop half of its initial value is

Oscillations

Solution:

The energy of the damped oscillator at any instant t is given by
$E=E_{0}e^{-bt/ m} \ldots\left(i\right)$
where $E_{0}$ is its initial energy and $b$ is the damping constant
At $t=t_{1/2}$, the energy drop to half of its initial value From Eq.(i), we get
$\frac{E_{0}}{2}=E_{0}e^{-bt_{1/ 2\,{}^/{ m}}} \frac{1}{2}=e^{-bt_{1 /2\,/^m}}$
Taking natural logarithm on both sides, we get
ln $\left(\frac{1}{2}\right)=-\frac{bt_{1 2}}{m}, t_{1 2}=-\frac{m\,ln\left(1 2\right)}{b} \ldots\left(ii\right)$
Here, ln $\left(1 2\right)=-0.693$
$b=40\,g\,s^{-1}, m=200\,g$
Substituting in Eq. $\left(ii\right)$, we get
$t_{ 1 /2}=\frac{0.693\times200\,g}{40\,g\,s^{-1}}=3.5\, s$