The displacement of a damped harmonic oscillator is given by $$x(t) = e^{-0.1t}\cos(10\pi t + \varphi)$$. Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:

We are given the displacement of the damped oscillator as

$$x(t)=e^{-0.1t}\cos(10\pi t+\varphi),$$

where $$t$$ is in seconds. In such an expression the quantity multiplying the cosine, here $$e^{-0.1t},$$ represents the amplitude envelope. At any instant

$$A(t)=e^{-0.1t}.$$

At the initial moment $$t=0,$$ the amplitude is

$$A(0)=e^{-0.1\,(0)}=e^{0}=1.$$

We are asked for the time $$t$$ at which the amplitude becomes one-half of this initial value, that is

$$A(t)=\frac12.$$

Setting the envelope equal to $$\dfrac12$$ gives the equation

$$e^{-0.1t}=\frac12.$$

To extract $$t$$ we take the natural logarithm (base $$e$$) of both sides. Recall the logarithm rule $$\ln(e^{y})=y$$. Applying this rule we obtain

$$\ln\!\bigl(e^{-0.1t}\bigr)=\ln\!\left(\frac12\right).$$

So

$$-0.1t=\ln\!\left(\frac12\right).$$

We know the numerical value $$\ln\!\left(\dfrac12\right)=-\ln 2\approx-0.6931.$$ Substituting this value, we have

$$-0.1t=-0.6931.$$

Now dividing both sides by $$-0.1$$ gives

$$t=\frac{-0.6931}{-0.1}=6.931.$$

Thus

$$t\approx6.93\ \text{s}.$$

When rounded to the nearest whole number, this is about $$7\ \text{s}$$. Looking at the options provided, the choice that is closest to $$6.93\ \text{s}$$ is $$7\ \text{s}$$.

Hence, the correct answer is Option D.