In damped oscillations, the amplitude is reduced to one-third of its initial value `a_(0)` at the end of 100 oscillations. When the oscillator completes 200 oscillations ,its amplitude must be

Options

A. `a_(0)//2`
B. `a_(0)//4`
C. `a_(0)//6`
D. `a_(0)//9`

Explanation

To solve the problem of damped oscillations, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Damped Oscillation Formula**:
The amplitude of a damped oscillator at any time \( t \) can be expressed as:
\[ A(t) = A_0 e^{-bt} \]
where \( A_0 \) is the initial amplitude, \( b \) is the damping coefficient, and \( t \) is the time.

2. **Amplitude After 100 Oscillations**:
According to the problem, the amplitude is reduced to one-third of its initial value after 100 oscillations. Thus, we can write:
\[ A(100T) = \frac{A_0}{3} \]
Here, \( T \) is the time period of one oscillation, so the total time for 100 oscillations is \( 100T \).

3. **Setting Up the Equation**:
Substituting into the damped oscillation formula:
\[ \frac{A_0}{3} = A_0 e^{-b(100T)} \]
Dividing both sides by \( A_0 \):
\[ \frac{1}{3} = e^{-b(100T)} \]

4. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{1}{3}\right) = -b(100T) \]
Rearranging this, we find:
\[ b(100T) = -\ln(3) \]

5. **Finding the Amplitude After 200 Oscillations**:
Now, we need to find the amplitude after 200 oscillations:
\[ A(200T) = A_0 e^{-b(200T)} \]
We can express \( b(200T) \) in terms of \( b(100T) \):
\[ b(200T) = 2b(100T) = 2(-\ln(3)) = -2\ln(3) \]

6. **Substituting Back into the Amplitude Formula**:
Thus, we have:
\[ A(200T) = A_0 e^{-2\ln(3)} \]
Using the property of exponents:
\[ e^{-2\ln(3)} = \frac{1}{(e^{\ln(3)})^2} = \frac{1}{3^2} = \frac{1}{9} \]
Therefore:
\[ A(200T) = A_0 \cdot \frac{1}{9} \]

7. **Final Result**:
The amplitude after 200 oscillations is:
\[ A(200T) = \frac{A_0}{9} \]

### Conclusion:
The amplitude after 200 oscillations is \(\frac{A_0}{9}\).