A pendulum with time period of `1s` is losing energy due to damping. At time its energy is `45 J`. If after completing `15` oscillations, its energy has become `15 J`. Its damping constant (in `s^(-1)`) is :-

Options

A. (a)`1/30In3`
B. (b)`1/15In3`
C. (c)2
D. (d)`1/2`

Explanation

To solve the problem of finding the damping constant of a pendulum losing energy, we can follow these steps:

### Step 1: Understand the Energy Loss in Damped Oscillations
The energy of a damped oscillator decreases exponentially over time. The relationship can be expressed as:
\[ E(t) = E_0 e^{-kt} \]
where:
- \( E(t) \) is the energy at time \( t \),
- \( E_0 \) is the initial energy,
- \( k \) is the damping constant,
- \( t \) is the time.

### Step 2: Identify Given Values
From the problem, we have:
- Initial energy \( E_0 = 45 \, J \)
- Energy after 15 oscillations \( E(15) = 15 \, J \)
- Time period \( T = 1 \, s \) (thus, \( t = 15 \, s \) for 15 oscillations)

### Step 3: Set Up the Equation
Using the energy loss formula:
\[ E(15) = E_0 e^{-k \cdot t} \]
Substituting the known values:
\[ 15 = 45 e^{-k \cdot 15} \]

### Step 4: Simplify the Equation
Divide both sides by 45:
\[ \frac{15}{45} = e^{-k \cdot 15} \]
This simplifies to:
\[ \frac{1}{3} = e^{-k \cdot 15} \]

### Step 5: Take the Natural Logarithm
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{1}{3}\right) = -k \cdot 15 \]

### Step 6: Solve for the Damping Constant \( k \)
Rearranging the equation:
\[ k = -\frac{\ln\left(\frac{1}{3}\right)}{15} \]
Using the property of logarithms, we can rewrite:
\[ \ln\left(\frac{1}{3}\right) = -\ln(3) \]
Thus:
\[ k = \frac{\ln(3)}{15} \]

### Step 7: Calculate the Value of \( k \)
Using the approximate value of \( \ln(3) \approx 1.0986 \):
\[ k \approx \frac{1.0986}{15} \approx 0.07324 \, s^{-1} \]

### Final Answer
The damping constant \( k \) is approximately \( 0.073 \, s^{-1} \).

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