Equation of motion for a particle performing damped harmonic oscillation is given as `x=e^(-0.1t) cos(10pit+phi)`. The time when amplitude will half of the initial is :

Options

A. 27
B. 4
C. 1
D. 7

Answer

To solve the problem, we need to determine the time when the amplitude of a damped harmonic oscillator is half of its initial value. The equation of motion given is:

\[ x(t) = e^{-0.1t} \cos(10\pi t + \phi) \]

### Step-by-Step Solution:

1. **Identify the Initial Amplitude**:
The initial amplitude \( A_0 \) can be identified from the equation. The amplitude of the oscillation is given by the exponential term:
\[ A(t) = A_0 e^{-0.1t} \]
Here, \( A_0 \) is the initial amplitude at \( t = 0 \).

2. **Set Up the Equation for Half Amplitude**:
We want to find the time \( t \) when the amplitude is half of the initial amplitude:
\[ A(t) = \frac{A_0}{2} \]
Substituting the expression for amplitude:
\[ A_0 e^{-0.1t} = \frac{A_0}{2} \]

3. **Cancel \( A_0 \)**:
Since \( A_0 \) is non-zero, we can divide both sides of the equation by \( A_0 \):
\[ e^{-0.1t} = \frac{1}{2} \]

4. **Take the Natural Logarithm**:
To solve for \( t \), we take the natural logarithm of both sides:
\[ \ln(e^{-0.1t}) = \ln\left(\frac{1}{2}\right) \]
This simplifies to:
\[ -0.1t = \ln\left(\frac{1}{2}\right) \]

5. **Solve for \( t \)**:
Rearranging the equation gives:
\[ t = \frac{\ln\left(\frac{1}{2}\right)}{-0.1} \]
We know that \( \ln\left(\frac{1}{2}\right) = -\ln(2) \), so:
\[ t = \frac{-\ln(2)}{-0.1} = \frac{\ln(2)}{0.1} \]

6. **Calculate \( t \)**:
The value of \( \ln(2) \) is approximately \( 0.693 \):
\[ t = \frac{0.693}{0.1} = 6.93 \text{ seconds} \]

7. **Round to the Nearest Whole Number**:
Rounding \( 6.93 \) seconds gives us approximately \( 7 \) seconds.

### Final Answer:
The time when the amplitude will be half of the initial amplitude is approximately **7 seconds**.