The amplitude of a S.H.O. reduces to 1/3 in first 20 secs. then in first 40 sec. its amplitude becomes -

Options

A. `1/3`
B. `1/9`
C. `1/27`
D. `1/sqrt3`

Explanation

To solve the problem step by step, we will analyze the given information about the simple harmonic oscillation (SHO) and apply the relevant equations.

### Step 1: Understand the equation of motion for SHO
The general equation for the displacement \( x \) of a simple harmonic oscillator is given by:
\[ x = x_0 e^{-\lambda t} \]
where:
- \( x_0 \) is the initial amplitude,
- \( \lambda \) is the damping constant,
- \( t \) is the time.

### Step 2: Set up the equation for the first condition
According to the problem, the amplitude reduces to \( \frac{1}{3} \) of its initial value after 20 seconds. Therefore, we can write:
\[ x(20) = \frac{x_0}{3} \]
Substituting this into the equation:
\[ \frac{x_0}{3} = x_0 e^{-\lambda \cdot 20} \]

### Step 3: Simplify the equation
We can cancel \( x_0 \) from both sides (assuming \( x_0 \neq 0 \)):
\[ \frac{1}{3} = e^{-\lambda \cdot 20} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{1}{3}\right) = -\lambda \cdot 20 \]
Thus, we can express \( \lambda \):
\[ \lambda = -\frac{\ln\left(\frac{1}{3}\right)}{20} \]

### Step 5: Set up the equation for the second condition
Now we need to find the amplitude after 40 seconds:
\[ x(40) = x_0 e^{-\lambda \cdot 40} \]
We can express \( -\lambda \cdot 40 \) as:
\[ -\lambda \cdot 40 = -\left(-\frac{\ln\left(\frac{1}{3}\right)}{20}\right) \cdot 40 = 2 \ln\left(\frac{1}{3}\right) \]
Thus, we can write:
\[ x(40) = x_0 e^{2 \ln\left(\frac{1}{3}\right)} \]

### Step 6: Simplify using properties of logarithms
Using the property of exponents:
\[ e^{\ln(a^b)} = a^b \]
we can simplify:
\[ x(40) = x_0 \left(\frac{1}{3}\right)^2 = x_0 \cdot \frac{1}{9} \]

### Step 7: Conclusion
Therefore, after 40 seconds, the amplitude becomes:
\[ x(40) = \frac{x_0}{9} \]

### Final Answer
The amplitude after 40 seconds is \( \frac{x_0}{9} \).

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