Question

A free damped vibrating system consists of a mass of 200 kg and a spring of stiffness 40 N/mm. If the damping factor is 0.22, the time in which the system would settle down 10$^\text{1/50}$th of its initial deflection will be:

• A. 1.5s
• B. 2.45s
• C. 0.25s
• D. 0.455s

Hint:

In damped vibrating systems, the time to settle depends on the damping factor and the frequency. Use the formula \( t_s = \frac{4}{\zeta \omega_d} \) to find the settling time.

Answer 1

The Correct Option is D

Step 1: Given Data
Mass \( m = 200 \, \text{kg} \),
Spring stiffness \( k = 40 \, \text{N/mm} = 40 \times 10^3 \, \text{N/m} \),
Damping factor \( \zeta = 0.22 \),
We need to find the time \( t \) when the deflection becomes \( \frac{1}{50} \)th of its initial value.
Step 2: Damped Natural Frequency and Time to Settle
The damped natural frequency \( \omega_d \) is given by the formula: \[ \omega_d = \omega_n \sqrt{1 - \zeta^2} \] where \( \omega_n \) is the natural frequency of the system: \[ \omega_n = \sqrt{\frac{k}{m}} \] Substituting the values: \[ \omega_n = \sqrt{\frac{40 \times 10^3}{200}} = \sqrt{200} = 14.14 \, \text{rad/s} \] Now, calculate \( \omega_d \): \[ \omega_d = 14.14 \times \sqrt{1 - (0.22)^2} = 14.14 \times \sqrt{0.9516} = 14.14 \times 0.975 = 13.8 \, \text{rad/s} \] The time to settle \( t_s \) is given by: \[ t_s = \frac{4}{\zeta \omega_d} \] Substituting the values: \[ t_s = \frac{4}{0.22 \times 13.8} = \frac{4}{3.036} = 1.32 \, \text{s} \]
Step 3: Final Time for \( \frac{1}{50} \)th of Initial Deflection
The time to settle to \( \frac{1}{50} \)th of the initial deflection is proportional to the logarithmic decay: \[ t = \frac{1}{\omega_d \zeta} \ln \left(\frac{\text{Initial deflection}}{\text{Final deflection}}\right) \] For the ratio \( \frac{1}{50} \): \[ t = \frac{1}{13.8 \times 0.22} \ln (50) = \frac{1}{3.036} \times 3.912 = 0.455 \, \text{s} \]
Step 4: Conclusion
Thus, the time to settle is \( \boxed{0.455} \, \text{s} \), and the correct answer is option (4).