The displacement of a damped harmonic oscillator is given by `x(t)=e^(-0.1t)cos(10pit+phi)`. Here t is in seconds
The time taken for its amplitude of vibration to drop to half of its initial is close to :

Options

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

Answer

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

### Step 1: Identify the initial amplitude
The displacement of the damped harmonic oscillator is given by:
\[ x(t) = e^{-0.1t} \cos(10\pi t + \phi) \]

Here, the amplitude of the oscillation is given by the term \( e^{-0.1t} \). At \( t = 0 \), the initial amplitude \( A_0 \) is:
\[ A_0 = e^{0} = 1 \]

### Step 2: Set up the equation for half the amplitude
We want to find the time \( t \) when the amplitude drops to half of its initial value:
\[ A(t) = \frac{A_0}{2} \]
Substituting the expression for amplitude, we have:
\[ e^{-0.1t} = \frac{1}{2} \]

### Step 3: Solve for \( t \)
To solve for \( t \), take the natural logarithm of both sides:
\[ -0.1t = \ln\left(\frac{1}{2}\right) \]
Now, rearranging gives:
\[ t = -\frac{\ln\left(\frac{1}{2}\right)}{0.1} \]

### Step 4: Calculate \( t \)
Using the property of logarithms, we know that:
\[ \ln\left(\frac{1}{2}\right) = -\ln(2) \]
Thus, we can rewrite the equation as:
\[ t = \frac{\ln(2)}{0.1} \]

Now, using the approximate value of \( \ln(2) \approx 0.693 \):
\[ t \approx \frac{0.693}{0.1} = 6.93 \text{ seconds} \]

### Final Answer
The time taken for the amplitude of vibration to drop to half of its initial value is approximately **6.93 seconds**.

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