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. `3^(2)`
B. `root(3)(2)`
C. `root(3)(3)`
D. `3^(3)`

Answer

To solve the problem, we need to analyze the behavior of a damped oscillator and use the provided information to find the value of \( n \).

### Step-by-Step Solution:

1. **Understanding the Damped Oscillator Equation**:
The amplitude \( A(t) \) of a damped oscillator at time \( t \) can be expressed as:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( A_0 \) is the initial amplitude and \( \lambda \) is the damping constant.

2. **Setting Up the First Condition**:
According to the problem, the amplitude becomes \( \frac{1}{3} A_0 \) after \( 2 \) seconds. We can set up the equation:
\[ A(2) = A_0 e^{-2\lambda} = \frac{1}{3} A_0 \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ e^{-2\lambda} = \frac{1}{3} \]
This is our **Equation (1)**.

3. **Setting Up the Second Condition**:
The problem states that after \( 6 \) seconds, the amplitude is \( \frac{1}{n} A_0 \). We can express this as:
\[ A(6) = A_0 e^{-6\lambda} = \frac{1}{n} A_0 \]
Again, dividing both sides by \( A_0 \):
\[ e^{-6\lambda} = \frac{1}{n} \]
This is our **Equation (2)**.

4. **Relating the Two Equations**:
From **Equation (1)**, we have:
\[ e^{-2\lambda} = \frac{1}{3} \]
We can express \( e^{-6\lambda} \) in terms of \( e^{-2\lambda} \):
\[ e^{-6\lambda} = (e^{-2\lambda})^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

5. **Substituting into Equation (2)**:
Now substituting \( e^{-6\lambda} \) into **Equation (2)**:
\[ \frac{1}{27} = \frac{1}{n} \]
This implies:
\[ n = 27 \]

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