A block of mass 200 g executing SHM under the influence of a spring of spring constant `k=90Nm^(-1)` and a damping constant `b=40gs^(-1)`. The time elapsed for its amplitude to drop to half of its initial value is (Given, ln `(1//2)=-0.693`)

Options

A. 7s
B. 9s
C. 4s
D. 11s

Explanation

To solve the problem step by step, we need to find the time elapsed for the amplitude of a damped oscillation to drop to half of its initial value.

### Step 1: Understand the formula for amplitude in damped oscillations
The amplitude of a damped oscillation at any time \( t \) is given by:
\[ A(t) = A_0 e^{-\frac{bt}{2m}} \]
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 block.

### Step 2: Set up the equation for the amplitude dropping to half
We want to find the time \( t \) when the amplitude drops to half its initial value:
\[ \frac{A_0}{2} = A_0 e^{-\frac{bt}{2m}} \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\frac{bt}{2m}} \]

### Step 3: Take the natural logarithm of both sides
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{1}{2}\right) = -\frac{bt}{2m} \]

### Step 4: Solve for \( t \)
Rearranging the equation to solve for \( t \):
\[ t = -\frac{2m \ln\left(\frac{1}{2}\right)}{b} \]

### Step 5: Substitute the values into the equation
We have:
- Mass \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (convert grams to kilograms),
- Damping constant \( b = 40 \, \text{g/s} = 0.04 \, \text{kg/s} \) (convert grams to kilograms),
- Given \( \ln\left(\frac{1}{2}\right) = -0.693 \).

Substituting these values into the equation:
\[ t = -\frac{2 \times 0.2 \times (-0.693)}{0.04} \]

### Step 6: Calculate the value of \( t \)
Calculating the numerator:
\[ 2 \times 0.2 \times 0.693 = 0.276 \]
Now, substituting this into the equation for \( t \):
\[ t = \frac{0.276}{0.04} = 6.9 \, \text{s} \]

### Final Answer
Thus, the time elapsed for the amplitude to drop to half of its initial value is approximately:
\[ t \approx 6.9 \, \text{s} \]