The amplitude (A) of damped oscillator becomes half in 5 minutes. The amplitude after next 10 minutes will be

Options

A. A
B. A/8
C. A/4
D. 4A

Answer

To solve the problem, we need to analyze the behavior of a damped oscillator, where the amplitude decreases exponentially over time.

### Step-by-Step Solution:

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

2. **Finding the Damping Constant \( \lambda \):**
We know that the amplitude becomes half after 5 minutes. Therefore, at \( t = 5 \) minutes:
\[ A(5) = \frac{A_0}{2} \]
Substituting into the equation:
\[ \frac{A_0}{2} = A_0 e^{-\lambda \cdot 5} \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-5\lambda} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -5\lambda \]
Thus, we can express \( \lambda \):
\[ \lambda = -\frac{\ln\left(\frac{1}{2}\right)}{5} = \frac{\ln(2)}{5} \]

3. **Calculating the Amplitude After 15 Minutes:**
Now we want to find the amplitude after 15 minutes:
\[ A(15) = A_0 e^{-\lambda \cdot 15} \]
Substituting \( \lambda \):
\[ A(15) = A_0 e^{-15 \cdot \frac{\ln(2)}{5}} = A_0 e^{-3\ln(2)} = A_0 (e^{\ln(2)})^{-3} = A_0 \left(\frac{1}{2}\right)^3 \]
Simplifying:
\[ A(15) = A_0 \cdot \frac{1}{8} = \frac{A_0}{8} \]

4. **Conclusion:**
The amplitude after the next 10 minutes (which is 15 minutes from the start) will be:
\[ A(15) = \frac{A_0}{8} \]

### Final Answer:
The amplitude after the next 10 minutes will be \( \frac{A_0}{8} \).

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