To solve the problem, we need to determine the time \( t \) at which the amplitude of a damped spring-mass system drops to \( \frac{1}{5} \) of its initial value. We will use the formula for the amplitude of a damped oscillator:
\[
A(t) = A_0 e^{-\frac{b}{2m} t}
\]
where:
- \( A(t) \) is the amplitude at time \( t \),
- \( A_0 \) is the initial amplitude,
- \( b \) is the damping constant,
- \( m \) is the mass of the system.
### Step 1: Set up the equation for the amplitude
We know that at time \( t \), the amplitude drops to \( \frac{1}{5} A_0 \). Therefore, we can write:
\[
\frac{1}{5} A_0 = A_0 e^{-\frac{b}{2m} t}
\]
### Step 2: Simplify the equation
We can divide both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[
\frac{1}{5} = e^{-\frac{b}{2m} t}
\]
### Step 3: Take the natural logarithm of both sides
Taking the natural logarithm gives us:
\[
\ln\left(\frac{1}{5}\right) = -\frac{b}{2m} t
\]
### Step 4: Use the property of logarithms
Using the property of logarithms, we can rewrite \( \ln\left(\frac{1}{5}\right) \) as:
\[
\ln\left(\frac{1}{5}\right) = -\ln(5)
\]
Thus, the equation becomes:
\[
-\ln(5) = -\frac{b}{2m} t
\]
### Step 5: Rearrange the equation for \( t \)
Cancelling the negative signs, we have:
\[
\ln(5) = \frac{b}{2m} t
\]
Now, rearranging for \( t \):
\[
t = \frac{2m \ln(5)}{b}
\]
### Step 6: Substitute the values
Given:
- Mass \( m = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Damping constant \( b = 10 \, \text{g/s} = 0.01 \, \text{kg/s} \) (since \( 10 \, \text{g/s} = 10 \times 10^{-3} \, \text{kg/s} \))
- \( \ln(5) = 1.61 \)
Substituting these values into the equation:
\[
t = \frac{2 \times 0.2 \, \text{kg} \times 1.61}{0.01 \, \text{kg/s}}
\]
### Step 7: Calculate \( t \)
Calculating the numerator:
\[
2 \times 0.2 \times 1.61 = 0.644 \, \text{kg}
\]
Now, substituting back into the equation for \( t \):
\[
t = \frac{0.644}{0.01} = 64.4 \, \text{s}
\]
### Step 8: Express \( t \) in terms of \( 2^x \)
We need to express \( t \) in the form \( 2^x \). Since \( 64 = 2^6 \), we can write:
\[
t \approx 64 \, \text{s} \implies t \approx 2^6 \, \text{s}
\]
### Conclusion
Thus, comparing \( t \) with \( 2^x \), we find:
\[
x = 6
\]
### Final Answer
The value of \( x \) is \( 6 \).
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