The amplitude of damped oscillator becomes `1/3` in `2s`. Its amplitude after `6s` is `1//n` times the original. The value of `n` is

Options

A. `2^(3)`
B. `3^(2)`
C. `3^(1//3)`
D. `3^(3)`

Explanation

To solve the problem, we need to analyze the behavior of a damped oscillator. The amplitude of a damped oscillator decreases exponentially over time, and we can use the formula for the amplitude of a damped oscillator:

\[ A(t) = A_0 e^{-\lambda t} \]

where:
- \( A(t) \) is the amplitude at time \( t \),
- \( A_0 \) is the initial amplitude,
- \( \lambda \) is the damping constant,
- \( t \) is the time.

### Step-by-Step Solution:

1. **Identify the given information**:
- At \( t = 2 \) seconds, the amplitude becomes \( \frac{1}{3} A_0 \).
- At \( t = 6 \) seconds, the amplitude is \( \frac{1}{n} A_0 \).

2. **Set up the equation for \( t = 2 \) seconds**:
\[ A(2) = A_0 e^{-\lambda \cdot 2} = \frac{1}{3} A_0 \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ e^{-2\lambda} = \frac{1}{3} \]

3. **Take the natural logarithm of both sides**:
\[ -2\lambda = \ln\left(\frac{1}{3}\right) \]
Thus,
\[ \lambda = -\frac{1}{2} \ln\left(\frac{1}{3}\right) \]

4. **Set up the equation for \( t = 6 \) seconds**:
\[ A(6) = A_0 e^{-\lambda \cdot 6} = \frac{1}{n} A_0 \]
Dividing both sides by \( A_0 \):
\[ e^{-6\lambda} = \frac{1}{n} \]

5. **Substituting the value of \( \lambda \)**:
\[ e^{-6\left(-\frac{1}{2} \ln\left(\frac{1}{3}\right)\right)} = \frac{1}{n} \]
Simplifying this gives:
\[ e^{3 \ln\left(\frac{1}{3}\right)} = \frac{1}{n} \]
Which can be rewritten as:
\[ \left(\frac{1}{3}\right)^3 = \frac{1}{n} \]

6. **Calculating \( n \)**:
\[ \frac{1}{27} = \frac{1}{n} \]
Therefore, we find:
\[ n = 27 \]

### Final Answer:
The value of \( n \) is \( 27 \).