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

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a damped harmonic oscillator with a frequency of 5 oscillations per second. The amplitude drops to half its value every 10 oscillations. We need to find the time it takes for the amplitude to drop to \( \frac{1}{1000} \) of its original value.

### Step 2: Calculate the time for 10 oscillations
Given the frequency \( f = 5 \) Hz, the time period \( T \) can be calculated as:
\[ T = \frac{1}{f} = \frac{1}{5} = 0.2 \text{ seconds} \]
The time for 10 oscillations is:
\[ t_{10} = 10 \times T = 10 \times 0.2 = 2 \text{ seconds} \]

### Step 3: Relate amplitude and time
The relationship between amplitude and time for a damped harmonic oscillator is given by:
\[ A(t) = A_0 e^{-\gamma t} \]
where \( A_0 \) is the initial amplitude and \( \gamma \) is the damping constant.

### Step 4: Find the damping constant \( \gamma \)
Since the amplitude drops to half its value after 10 oscillations (2 seconds), we have:
\[ \frac{A_0}{2} = A_0 e^{-\gamma \cdot 2} \]
Dividing both sides by \( A_0 \):
\[ \frac{1}{2} = e^{-2\gamma} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -2\gamma \]
This simplifies to:
\[ - \ln(2) = -2\gamma \implies \gamma = \frac{\ln(2)}{2} \]

### Step 5: Set up the equation for \( \frac{1}{1000} \) of the original amplitude
We want to find the time \( t \) when the amplitude is \( \frac{A_0}{1000} \):
\[ \frac{A_0}{1000} = A_0 e^{-\gamma t} \]
Dividing both sides by \( A_0 \):
\[ \frac{1}{1000} = e^{-\gamma t} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{1000}\right) = -\gamma t \]
This simplifies to:
\[ -3\ln(10) = -\gamma t \implies t = \frac{3\ln(10)}{\gamma} \]

### Step 6: Substitute \( \gamma \) into the equation
Substituting \( \gamma = \frac{\ln(2)}{2} \):
\[ t = \frac{3\ln(10)}{\frac{\ln(2)}{2}} = \frac{6\ln(10)}{\ln(2)} \]

### Step 7: Calculate the numerical value
Using approximate values \( \ln(10) \approx 2.303 \) and \( \ln(2) \approx 0.693 \):
\[ t \approx \frac{6 \times 2.303}{0.693} \approx \frac{13.818}{0.693} \approx 19.94 \text{ seconds} \]
Thus, the time it will take to drop to \( \frac{1}{1000} \) of the original amplitude is approximately 20 seconds.

### Final Answer:
The time it takes to drop to \( \frac{1}{1000} \) of the original amplitude is close to **20 seconds**.

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