Question

In a single degree of freedom vibrating system with only viscous damping, the critical damping coefficient is 350 N s/m and the damping coefficient is 35 N s/m.
The logarithmic decrement of the vibrating system is

• A. 0.63
• B. 1.26
• C. 0.32
• D. 1.89

Hint:

Get the damping ratio first from c over c_c, then plug it into the logarithmic decrement formula.

Answer 1

The Correct Option is A

Step 1: Get the damping ratio from the given data.
Damping ratio $\zeta$ equals the given damping coefficient over the critical damping coefficient, since critical damping is the value at which $\zeta = 1$. Here $\zeta = 35/350 = 0.1$, so the system is lightly underdamped.

Step 2: Recall what logarithmic decrement means.
Logarithmic decrement $\delta$ is the natural log of the ratio of two successive peak amplitudes in free vibration, and it links to $\zeta$ through $\delta = \dfrac{2\pi\zeta}{\sqrt{1-\zeta^2}}$.

Step 3: Plug in numbers and simplify.
$\sqrt{1-\zeta^2} = \sqrt{1-0.01} = \sqrt{0.99} \approx 0.995$. So $\delta = \dfrac{2\pi \times 0.1}{0.995} = \dfrac{0.6283}{0.995} \approx 0.632$.

Final Answer:
The closest match among the given choices is 0.63. \[ \boxed{\delta \approx 0.63} \]