A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to `1/1000` of the original amplitude is close to:

Options

A. 100s
B. 10s
C. 50s
D. 20s

Explanation

To solve the problem of a damped harmonic oscillator, we can follow these steps:

### Step 1: Understand the given parameters
- The frequency of oscillation is given as \( f = 5 \) oscillations per second.
- The amplitude drops to half its value after every 10 oscillations.

### Step 2: Calculate the time for 10 oscillations
Since the frequency is \( 5 \) oscillations per second, the time for \( 10 \) oscillations can be calculated as:
\[ t_{10} = \frac{10 \text{ oscillations}}{5 \text{ oscillations/second}} = 2 \text{ seconds} \]

### Step 3: Relate amplitude with time
The amplitude of a damped harmonic oscillator can be expressed as:
\[ A(t) = A_0 e^{-\frac{bt}{2m}} \]
Where:
- \( A_0 \) is the initial amplitude,
- \( b \) is the damping constant,
- \( m \) is the mass of the oscillator,
- \( t \) is time.

### Step 4: Set up the equation for half amplitude
After \( 2 \) seconds, the amplitude drops to half:
\[ \frac{A_0}{2} = A_0 e^{-\frac{b \cdot 2}{2m}} \]
Dividing both sides by \( A_0 \) gives:
\[ \frac{1}{2} = e^{-\frac{b \cdot 2}{2m}} \]

### Step 5: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\frac{b \cdot 2}{2m} \]
This simplifies to:
\[ -\ln(2) = -\frac{b}{m} \]
Thus, we find:
\[ \frac{b}{m} = \ln(2) \]

### Step 6: Set up the equation for \( \frac{A_0}{1000} \)
We want to find the time \( t_0 \) when the amplitude drops to \( \frac{A_0}{1000} \):
\[ \frac{A_0}{1000} = A_0 e^{-\frac{bt_0}{2m}} \]
Dividing both sides by \( A_0 \):
\[ \frac{1}{1000} = e^{-\frac{bt_0}{2m}} \]

### Step 7: Take the natural logarithm again
Taking the natural logarithm:
\[ \ln\left(\frac{1}{1000}\right) = -\frac{bt_0}{2m} \]
This can be rewritten as:
\[ -\ln(1000) = -\frac{bt_0}{2m} \]
Thus:
\[ \frac{bt_0}{2m} = \ln(1000) \]

### Step 8: Substitute \( \frac{b}{m} \) into the equation
Now substituting \( \frac{b}{m} = \ln(2) \):
\[ \ln(2) \cdot t_0 = 2 \ln(1000) \]
Solving for \( t_0 \):
\[ t_0 = \frac{2 \ln(1000)}{\ln(2)} \]

### Step 9: Calculate \( \ln(1000) \) and \( \ln(2) \)
We know:
\[ \ln(1000) = \ln(10^3) = 3 \ln(10) \approx 3 \times 2.303 = 6.909 \]
And:
\[ \ln(2) \approx 0.693 \]

### Step 10: Substitute and calculate \( t_0 \)
Substituting these values:
\[ t_0 = \frac{2 \times 6.909}{0.693} \approx \frac{13.818}{0.693} \approx 19.93 \text{ seconds} \]

### Conclusion
Thus, the time it will take for the amplitude to drop to \( \frac{1}{1000} \) of the original amplitude is approximately \( 20 \) seconds.