Calculating Viscous Damping Factor from Logarithmic Decrement

This problem requires calculating the viscous damping factor ($\zeta$) for a vibratory system. We are given the logarithmic decrement ($\delta$) measured over a specific number of cycles (five cycles) and its value is 8.11.

Understanding Key Concepts in Vibration

Before diving into the calculation, let's clarify the terms:

• Viscous Damping Factor ($\zeta$): This is a dimensionless parameter representing the level of damping in a system relative to the critical damping level. It's often called the damping ratio.
• Logarithmic Decrement ($\delta$): This parameter quantifies the rate at which the amplitude of free vibrations decays in a damped system. It is defined as the natural logarithm of the ratio of two successive amplitudes, separated by one cycle.

Calculating Logarithmic Decrement Per Cycle

The problem states the logarithmic decrement value *over five cycles* is 8.11. This usually represents the total reduction in the logarithm of the amplitude from the beginning (cycle 0) to the end of the fifth cycle (cycle 5). The amplitude ($x_N$) after $N$ cycles relates to the initial amplitude ($x_0$) by the formula $x_N = x_0 e^{-N\delta}$.

The logarithmic decrement over $N$ cycles is therefore $N\delta$. Given that the value over 5 cycles is 8.11, we have:

$$ 5\delta = 8.11 $$

To find the logarithmic decrement per cycle ($\delta$), we divide the total decrement by the number of cycles:

$$ \delta = \frac{8.11}{5} $$

$$ \delta = 1.622 $$

Relating Logarithmic Decrement to Viscous Damping Factor

The relationship between the logarithmic decrement ($\delta$) and the viscous damping factor ($\zeta$) is given by the formula:

$$ \delta = \frac{2 \pi \zeta}{\sqrt{1 - \zeta^2}} $$

This formula is accurate for all levels of damping (where $\zeta < 1$).

Step-by-Step Calculation of Viscous Damping Factor ($\zeta$)

1. Isolate $\zeta$ in the formula: We need to rearrange the equation $\delta = \frac{2 \pi \zeta}{\sqrt{1 - \zeta^2}}$ to solve for $\zeta$. First, square both sides:

   $$ \delta^2 = \frac{4 \pi^2 \zeta^2}{1 - \zeta^2} $$

   Multiply both sides by $(1 - \zeta^2)$:

   $$ \delta^2 (1 - \zeta^2) = 4 \pi^2 \zeta^2 $$

   Distribute $\delta^2$:

   $$ \delta^2 - \delta^2 \zeta^2 = 4 \pi^2 \zeta^2 $$

   Move terms involving $\zeta^2$ to one side:

   $$ \delta^2 = 4 \pi^2 \zeta^2 + \delta^2 \zeta^2 $$

   Factor out $\zeta^2$:

   $$ \delta^2 = \zeta^2 (4 \pi^2 + \delta^2) $$

   Solve for $\zeta^2$:

   $$ \zeta^2 = \frac{\delta^2}{4 \pi^2 + \delta^2} $$

   Finally, take the square root:

   $$ \zeta = \frac{\delta}{\sqrt{4 \pi^2 + \delta^2}} $$

2. Substitute the value of $\delta$: We calculated $\delta = 1.622$. Substitute this into the formula for $\zeta$. We use the value of $\pi \approx 3.14159$.

   $$ \zeta = \frac{1.622}{\sqrt{4 \pi^2 + (1.622)^2}} $$

3. Perform the calculation: Calculate the terms inside the square root: $4 \pi^2 \approx 4 \times (3.14159)^2 \approx 4 \times 9.86960 = 39.4784$ $(1.622)^2 \approx 2.630884$ Add them: $39.4784 + 2.630884 = 42.109284$ Take the square root: $\sqrt{42.109284} \approx 6.48917$ Now, divide:

   $$ \zeta = \frac{1.622}{6.48917} $$

   $$ \zeta \approx 0.25004 $$

4. Convert to percentage: The viscous damping factor $\zeta$ is often expressed as a percentage.

   $$ \zeta \approx 0.25004 \times 100\% $$

   $$ \zeta \approx 25.004\% $$

Rounding to the nearest whole percentage, the viscous damping factor is 25%.

Conclusion

By calculating the logarithmic decrement per cycle ($\delta = 1.622$) and using the relationship $\zeta = \frac{\delta}{\sqrt{4 \pi^2 + \delta^2}}$, we find the viscous damping factor ($\zeta$) to be approximately 25%. This corresponds to the second option provided.