The amplitude of a damped oscillator decreases to `0.9` times ist oringinal magnitude in `5s`, In anothet `10s` it will decrease to α times its original magnitude, where α equals to .

Options

A. 0.81
B. 0.729
C. 0.6
D. 0.7

Answer

To solve the problem, we need to find the value of α, which represents the amplitude of the damped oscillator after 15 seconds, given that the amplitude decreases to 0.9 times its original value in 5 seconds.

### Step-by-Step Solution:

1. **Understanding the Damped Oscillator Equation**:
The amplitude \( A(t) \) of a damped oscillator can be expressed as:
\[ A(t) = A_0 e^{-\frac{B}{2m} t} \]
where:
- \( A_0 \) is the initial amplitude,
- \( B \) is the damping coefficient,
- \( m \) is the mass of the oscillator,
- \( t \) is the time.

2. **Setting Up the Initial Condition**:
From the problem, we know that after 5 seconds, the amplitude decreases to 0.9 times its original magnitude:
\[ A(5) = 0.9 A_0 \]
Substituting into the equation:
\[ 0.9 A_0 = A_0 e^{-\frac{B}{2m} \cdot 5} \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ 0.9 = e^{-\frac{B}{2m} \cdot 5} \]

3. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln(0.9) = -\frac{B}{2m} \cdot 5 \]
Rearranging gives:
\[ \frac{B}{2m} = -\frac{\ln(0.9)}{5} \]

4. **Finding the Amplitude After 15 Seconds**:
We need to find the amplitude after 15 seconds:
\[ A(15) = A_0 e^{-\frac{B}{2m} \cdot 15} \]
Substituting \( \frac{B}{2m} \) from the previous step:
\[ A(15) = A_0 e^{-15 \left(-\frac{\ln(0.9)}{5}\right)} = A_0 e^{3 \ln(0.9)} \]
This simplifies to:
\[ A(15) = A_0 (e^{\ln(0.9)})^3 = A_0 (0.9)^3 \]

5. **Calculating \( \alpha \)**:
Now we calculate \( (0.9)^3 \):
\[ (0.9)^3 = 0.729 \]
Therefore, \( \alpha = 0.729 \).

### Final Answer:
\[ \alpha = 0.729 \]