Amplitude of a damped oscillator decreases up to 0.6 times of its initial value in 5 seconds. In next 10 seconds, it decreases upto `'alpha'` times of its intial value where `'alpha'` is equal to ?

Options

A. 0.729
B. 0.216
C. 0.7
D. 0.81

Answer

To solve the problem, we need to analyze the behavior of a damped oscillator and how its amplitude decreases over time.

### Step-by-Step Solution:

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

2. **Amplitude after 5 seconds**:
According to the problem, the amplitude decreases to 0.6 times its initial value in 5 seconds. Therefore, we can write:
\[ A(5) = 0.6 A_0 \]
Substituting into the amplitude equation:
\[ 0.6 A_0 = A_0 e^{-\frac{b}{2m} \cdot 5} \]
Dividing both sides by \( A_0 \):
\[ 0.6 = e^{-\frac{5b}{2m}} \]

3. **Taking the natural logarithm**:
To solve for \( \frac{b}{2m} \), we take the natural logarithm of both sides:
\[ \ln(0.6) = -\frac{5b}{2m} \]
Rearranging gives:
\[ \frac{b}{2m} = -\frac{\ln(0.6)}{5} \]

4. **Amplitude after 15 seconds**:
Now, we need to find the amplitude after a total of 15 seconds (5 seconds + 10 seconds). The amplitude at this time can be expressed as:
\[ A(15) = A_0 e^{-\frac{b}{2m} \cdot 15} \]
We want to express this in terms of \( \alpha \):
\[ A(15) = \alpha A_0 \]
Thus:
\[ \alpha A_0 = A_0 e^{-\frac{b}{2m} \cdot 15} \]
Dividing both sides by \( A_0 \):
\[ \alpha = e^{-\frac{15b}{2m}} \]

5. **Substituting the value of \( \frac{b}{2m} \)**:
We can substitute our earlier expression for \( \frac{b}{2m} \):
\[ \alpha = e^{-15 \left(-\frac{\ln(0.6)}{5}\right)} = e^{3 \ln(0.6)} \]
This simplifies to:
\[ \alpha = (0.6)^3 \]

6. **Calculating \( \alpha \)**:
Now we compute \( (0.6)^3 \):
\[ \alpha = 0.216 \]

### Final Answer:
Thus, the value of \( \alpha \) is:
\[ \alpha = 0.216 \]