In damped oscillations damping froce is directly proportional to speed of ocillatior .If amplitude becomes half to its maximum value is 1 s, then after 2 s amplitude will be (`A_(0)`- initial amplitude)

Options

A. `(1)/(4)A_(0)`
B. `(1)/(2)A_(0)`
C. `A_(0)`
D. `(sqrt(3)A_(0))/(2)`

Explanation

To solve the problem step by step, we will follow the concepts of damped oscillations and the given conditions.

### Step 1: Understand the formula for amplitude in damped oscillations
In damped oscillations, the amplitude \( A(t) \) at any time \( t \) is given by the formula:
\[ A(t) = A_0 e^{-\gamma t} \]
where:
- \( A_0 \) is the initial amplitude,
- \( \gamma \) is the damping coefficient,
- \( t \) is the time.

### Step 2: Set up the equation for the first condition
We are given that the amplitude becomes half of its maximum value in 1 second. This means:
\[ A(1) = \frac{A_0}{2} \]
Substituting into the amplitude formula:
\[ \frac{A_0}{2} = A_0 e^{-\gamma \cdot 1} \]

### Step 3: Simplify the equation
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\gamma} \]
Taking the natural logarithm of both sides:
\[ -\gamma = \ln\left(\frac{1}{2}\right) \]
This implies:
\[ \gamma = -\ln\left(\frac{1}{2}\right) = \ln(2) \]

### Step 4: Set up the equation for the second condition
Now we need to find the amplitude at \( t = 2 \) seconds:
\[ A(2) = A_0 e^{-\gamma \cdot 2} \]
Substituting \( \gamma = \ln(2) \):
\[ A(2) = A_0 e^{-2 \ln(2)} \]

### Step 5: Simplify the expression for amplitude at \( t = 2 \)
Using the property of exponents:
\[ e^{-2 \ln(2)} = \left(e^{\ln(2)}\right)^{-2} = 2^{-2} = \frac{1}{4} \]
Thus:
\[ A(2) = A_0 \cdot \frac{1}{4} = \frac{A_0}{4} \]

### Final Answer
The amplitude after 2 seconds will be:
\[ A(2) = \frac{A_0}{4} \]

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