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
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
The logarithmic decrement of the vibrating system is
Show Hint
Get the damping ratio first from c over c_c, then plug it into the logarithmic decrement formula.
Updated On: Jul 27, 2026
- 0.63
- 1.26
- 0.32
- 1.89
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Find the damping ratio.
The damping ratio is the actual damping coefficient divided by the critical damping coefficient: \(\zeta = \dfrac{c}{c_c} = \dfrac{35}{350} = 0.1\).
Step 2: Use the logarithmic decrement formula.
For an underdamped single degree of freedom system, the logarithmic decrement, the natural log of the ratio of two successive amplitude peaks, is \(\delta = \dfrac{2\pi\zeta}{\sqrt{1-\zeta^2}}\).
Step 3: Substitute and calculate.
\(\delta = \dfrac{2\pi(0.1)}{\sqrt{1-0.01}} = \dfrac{0.6283}{\sqrt{0.99}} = \dfrac{0.6283}{0.995} = 0.6316\).
Final Answer:
Rounding gives a logarithmic decrement close to 0.63. \[ \boxed{\delta \approx 0.63} \]
The damping ratio is the actual damping coefficient divided by the critical damping coefficient: \(\zeta = \dfrac{c}{c_c} = \dfrac{35}{350} = 0.1\).
Step 2: Use the logarithmic decrement formula.
For an underdamped single degree of freedom system, the logarithmic decrement, the natural log of the ratio of two successive amplitude peaks, is \(\delta = \dfrac{2\pi\zeta}{\sqrt{1-\zeta^2}}\).
Step 3: Substitute and calculate.
\(\delta = \dfrac{2\pi(0.1)}{\sqrt{1-0.01}} = \dfrac{0.6283}{\sqrt{0.99}} = \dfrac{0.6283}{0.995} = 0.6316\).
Final Answer:
Rounding gives a logarithmic decrement close to 0.63. \[ \boxed{\delta \approx 0.63} \]
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