If the natural frequency of oscillation ω_n = 13 rad/sec and damping ratio ξ is 0.8 then find the peak time.

Question

Question

This question was previously asked in

DFCCIL Executive S&T 2016 Official Paper

Answer 1

Concept:

Time-domain specification (or) transient response parameters:

Peak Time (tp): It is the time taken by the response to reach the maximum value.

\({\left. {\frac{{dc\left( t \right)}}{{dt}}} \right|_{t = {t_p}}} = 0,{\text{}}{t_p} = \frac{\pi }{{{ω _d}}}\)

\(As \ \omega_d=\omega_n\sqrt{ 1-\xi^2}\)

\(=\frac{\pi }{{{ω _n}√ {1 - {\zeta ^2}} }}\)

Where,

ξ = damping ratio

ω_n = natural frequency

ωd = damped frequency

Calculation:

Given:

ωn = 13 rad/sec

ξ = 0.8

\(\omega_d=13\sqrt{ 1-0.8^2}\)

ωd = 7.8 rad/sec.

\(t_p=\frac{\pi}{7.8}=0.4 \;sec\)

Hence option(4) is the correct answer.