The displacement of a damped harmonic oscillator is given by
`x(t)-e^(-0.1t) cos(10pit+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 :

Options

A. 4 s
B. 7 s
C. 13 s
D. 27 s

Answer

To solve the problem of finding the time taken for the amplitude of a damped harmonic oscillator to drop to half of its initial value, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equation**: The displacement of the damped harmonic oscillator is given by:
\[ x(t) = e^{-0.1t} \cos(10\pi t + \varphi) \]
Here, \(e^{-0.1t}\) represents the amplitude of the oscillation.

2. **Understand the amplitude**: The amplitude of the oscillation at time \(t\) is given by the term \(A(t) = e^{-0.1t}\).

3. **Determine the initial amplitude**: At \(t = 0\), the initial amplitude \(A(0)\) is:
\[ A(0) = e^{-0.1 \cdot 0} = e^0 = 1 \]

4. **Set up the equation for half amplitude**: We want to find the time \(t_{1/2}\) when the amplitude drops to half of its initial value:
\[ A(t_{1/2}) = \frac{1}{2} A(0) = \frac{1}{2} \]
Therefore, we have:
\[ e^{-0.1t_{1/2}} = \frac{1}{2} \]

5. **Take the natural logarithm of both sides**: To solve for \(t_{1/2}\), we take the natural logarithm:
\[ -0.1t_{1/2} = \ln\left(\frac{1}{2}\right) \]

6. **Solve for \(t_{1/2}\)**: Rearranging gives:
\[ t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-0.1} \]
Since \(\ln\left(\frac{1}{2}\right) = -\ln(2)\), we can write:
\[ t_{1/2} = \frac{-\ln(2)}{-0.1} = \frac{\ln(2)}{0.1} \]

7. **Calculate \(\ln(2)\)**: The value of \(\ln(2)\) is approximately \(0.693\). Thus:
\[ t_{1/2} = \frac{0.693}{0.1} = 6.93 \text{ seconds} \]

8. **Conclusion**: The time taken for the amplitude of vibration to drop to half of its initial value is approximately \(6.93\) seconds, which is close to \(7\) seconds.

### Final Answer:
The time taken for its amplitude of vibration to drop to half of its initial value is approximately \(7\) seconds.