A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 N m$$^{-1}$$ and oscillates in a damping medium of damping constant $$10^{-2}$$ kg s$$^{-1}$$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to-

For a damped harmonic oscillator, the equation of motion is $$m\,\ddot x + b\,\dot x + k\,x = 0$$, where $$m$$ is the mass, $$b$$ is the damping constant and $$k$$ is the spring (force) constant.

We identify, from the data given,

$$m = 0.1\ \text{kg}, \qquad k = 640\ \text{N\,m}^{-1}, \qquad b = 10^{-2}\ \text{kg\,s}^{-1}.$$

The quantity that governs the rate at which the amplitude (and hence the energy) decays is the damping coefficient $$\beta$$, defined by the relation

$$\beta = \frac{b}{2m}.$$

Substituting the numerical values, we get

$$\beta \;=\; \frac{10^{-2}}{2 \times 0.1} \;=\; \frac{10^{-2}}{0.2} \;=\; 5 \times 10^{-2}\ \text{s}^{-1} \;=\; 0.05\ \text{s}^{-1}.$$

In a lightly damped oscillator the total mechanical energy $$E(t)$$ diminishes exponentially with time according to the well-known relation

$$E(t) = E_0\,e^{-2\beta t},$$

where $$E_0$$ is the initial energy and $$t$$ is the elapsed time.

The question asks for the time $$t$$ at which the energy becomes exactly one-half of its initial value, so we set

$$\frac{E(t)}{E_0} = \frac12.$$

Using the exponential law, this requirement reads

$$e^{-2\beta t} = \frac12.$$

We now take natural logarithms of both sides. Recalling that the natural logarithm of $$\tfrac12$$ is $$-\ln 2$$, we obtain

$$-2\beta t = \ln\!\left(\frac12\right) = -\ln 2.$$

Dividing throughout by $$-2\beta$$ gives

$$t = \frac{\ln 2}{2\beta}.$$

We already know $$\beta = 0.05\ \text{s}^{-1}$$, so substituting we have

$$t = \frac{\ln 2}{2 \times 0.05} = \frac{0.693}{0.10} = 6.93\ \text{s}.$$

This numerical value is closest to $$7\ \text{s}$$ among the options supplied.

Hence, the correct answer is Option D.