The amplitude of a damped oscillator becomes `(1)/(27)^(th)` of its initial value after `6` minutes. Its amplitude after `2` minutes is

Options

A. `(A_(0))/(3)`
B. `(A_(0))/(9)`
C. `(A_(0))/(54)`
D. `(A_(0))/(81)`

Explanation

To solve the problem, we need to find the amplitude of a damped oscillator after 2 minutes, given that its amplitude becomes \( \frac{1}{27} \) of its initial value after 6 minutes.

### Step-by-Step Solution:

1. **Understanding the Damped Oscillator Formula**:
The amplitude \( A \) of a damped oscillator at any time \( t \) is given by the formula:
\[ A = A_0 e^{-bt} \]
where:
- \( A_0 \) is the initial amplitude,
- \( b \) is the damping constant,
- \( t \) is the time.

2. **Setting Up the Equation for 6 Minutes**:
According to the problem, after 6 minutes, the amplitude becomes \( \frac{1}{27} A_0 \):
\[ A(6) = A_0 e^{-b \cdot 6} = \frac{1}{27} A_0 \]
Dividing both sides by \( A_0 \) gives:
\[ e^{-b \cdot 6} = \frac{1}{27} \]

3. **Taking the Natural Logarithm**:
To solve for \( b \), we take the natural logarithm of both sides:
\[ -b \cdot 6 = \ln\left(\frac{1}{27}\right) \]
This simplifies to:
\[ b = -\frac{\ln\left(\frac{1}{27}\right)}{6} \]

4. **Finding the Amplitude After 2 Minutes**:
Now, we need to find the amplitude after 2 minutes:
\[ A(2) = A_0 e^{-b \cdot 2} \]
Substituting \( b \) from the previous step:
\[ A(2) = A_0 e^{-(-\frac{\ln\left(\frac{1}{27}\right)}{6}) \cdot 2} \]
This simplifies to:
\[ A(2) = A_0 e^{\frac{2}{6} \ln\left(\frac{1}{27}\right)} = A_0 e^{\frac{1}{3} \ln\left(\frac{1}{27}\right)} \]

5. **Using Properties of Exponents**:
We can rewrite the expression using properties of exponents:
\[ A(2) = A_0 \left(\frac{1}{27}\right)^{\frac{1}{3}} = A_0 \cdot \frac{1}{3} \]

6. **Final Answer**:
Therefore, the amplitude after 2 minutes is:
\[ A(2) = \frac{A_0}{3} \]