Question

A pendulum of time period one second is losing its mechanical energy due to damping. Its mechanical energy at time $ t = 0 $ is 45 J. After completing 15 oscillations, its mechanical energy is 15 J. The ratio of the damping constant and the mass of the object making damped oscillations is:

• A. \( \frac{1}{5} \log_e 3 \, \text{s}^{-1} \)
• B. \( \frac{1}{10} \log_e 3 \, \text{s}^{-1} \)
• C. \( \frac{1}{15} \log_e 3 \, \text{s}^{-1} \)
• D. \( \frac{1}{20} \log_e 3 \, \text{s}^{-1} \)

Hint:

When solving damped oscillation problems, remember that mechanical energy decreases exponentially. Using the logarithmic form will help to find the damping constant.

Answer 1

The Correct Option is C

The mechanical energy of a damped oscillating system decreases exponentially with time. The equation that describes this decay is: \[ E(t) = E_0 e^{-2 \gamma t} \] Where:
\( E(t) \) is the mechanical energy after time \( t \),
\( E_0 \) is the initial mechanical energy,
\( \gamma \) is the damping constant, and
\( t \) is the time elapsed.
Step 1: Substitute known values into the equation. Given that:
The initial mechanical energy \( E_0 = 45 \, \text{J} \) at time \( t = 0 \),
After 15 oscillations, the mechanical energy is \( E(15) = 15 \, \text{J} \), and
The time period of the pendulum is 1 second, so after 15 oscillations, \( t = 15 \, \text{s} \). Substitute these values into the equation: \[ E(15) = E_0 e^{-2 \gamma \cdot 15} \] \[ 15 = 45 e^{-30 \gamma} \] Step 2: Simplify the equation. Divide both sides of the equation by 45: \[ \frac{1}{3} = e^{-30 \gamma} \] Step 3: Take the natural logarithm of both sides. Apply \( \log_e \) to both sides: \[ \log_e \frac{1}{3} = -30 \gamma \] We know that \( \log_e \frac{1}{3} = -\log_e 3 \), so the equation becomes: \[ -\log_e 3 = -30 \gamma \] Step 4: Solve for \( \gamma \). Simplify: \[ \gamma = \frac{1}{30} \log_e 3 \] Step 5: Calculate the ratio of the damping constant to the mass.
Since we are asked to find the ratio of the damping constant \( \gamma \) and the mass \( m \), assuming \( m = 1 \), the ratio is: \[ \frac{\gamma}{m} = \frac{1}{30} \log_e 3 \] Thus, the correct answer is: \[ \boxed{\frac{1}{15} \log_e 3 \, \text{s}^{-1}} \]