Logarithmic decrement of a damped single degree of freedom system is \delta. If stiffness of the spring is doubled and mass is made half, then logarithmic decrement of the new system will be equal to
- A. 1/2 \delta
- B. \delta
- C. 2 \delta
- D. 1/2 \delta
Answer: B. \delta
Explanation: Log decrement \delta = 2\pi \zeta / $\sqrt{1-\zeta²}$. Damping ratio \zeta = c / 2\sqrt{mk}. New k' = 2k, m' = m/2. \sqrt{m'k'} = \sqrt{(m/2)(2k)} = \sqrt{mk}. So \zeta remains same. Thus \delta remains same.