English
Hindi

Question

Download Solution PDF

A second order control system has a damping ratio as 0.6 and natural frequency of oscillations as 11 rad/sec. What will be the Damped frequency of oscillation?

This question was previously asked in
DFCCIL Executive S&T 2016 Official Paper
Download PDF Attempt in App
View all DFCCIL Executive Papers >
  1. 2.6 rad/sec
  2. 8.8 rad/sec
  3. 6.9 rad/sec
  4. 5.6 rad/sec

Answer (Detailed Solution Below)

Option 2 : 8.8 rad/sec
Free Tests
View all Free tests >
Free
Intelligence Test Battery Practice Set 1 (Easy to Moderate)
35 Qs. 35 Marks 10 Mins
Start Now

Detailed Solution

Download Solution PDF

Concept:

Damped natural frequency of a second order system is given by

Where,

ωn = Natural frequency

ζ = Damping ratio

Calculation:

Given-

ωn = 11 rad/sec

ζ = 0.6

Now, natural damped frequency can be calculated as

ωd = 11 x 0.8

ωd = 8.8 rad/sec  

Download Solution PDF
Latest DFCCIL Executive Updates

Last updated on Jul 18, 2025

-> DFCCIL Executive Answer Key 2025 has been released at its official website. Candidates can calculate their marks, estimate their scorecard with log in ID and Password.

-> The Objection window will remain open from 18th July to 22nd July 2025. Candidates can raise their objection with qid no. and proof. 

-> The DFCCIL Rank Predictor 2025 is now available.

-> DFCCIL Executive Exam Analysis 2025 is live now. candidates can check level of exam, and estimate their score by analysing CBT 1 exam here.

-> DFCCIL Executive city intimation slip has been released. 

->  DFCCIL Executive Recruitment 2025 Correction window is open from 31st March, to 4th April 2025.

-> Candidates can make corrections in their application form, if they have made any mistake. There will be 100/- fee for correction in form.

-> DFCCIL Executive Recruitment 2025 application deadline has been extended.

-> Eligible and Interested candidates had applied from 18th January 2025 to 22nd March 2025.  

-> A total of 175 Vacancies have been announced for multiple posts like Executive (Civil, Electrical, Signal and Telecommunication).

-> Candidates who will get a successful selection under the DFCCIL Recruitment 2025 for the Executive selection process will get a salary range between Rs. 30,000 to Rs. 1,20,000.

-> Candidates must refer to the DFCCIL Executive Previous Year Papers to prepare well for the exam.

Exams Tests Super Practice

More Time Domain Specifications Questions

Q1.In the circuit shown switch is closed at t = 0. The time constant of the circuit is :
Q2.Consider a causal second-order system with the transfer function \(G(s)=\frac{1}{s^2+2 s+1}\) with a unit-step \(R(s)=\frac{1}{s}\) as an input. Let c(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value \(\displaystyle \lim _{t \rightarrow \infty} c(t)\), rounded off to two decimal places, is
Q3.The nature of the response shown below is:
Q4.When the damping ratio ζ (zeta) lies in between 0 < ζ < 1, then the circuit classification comes under:
Q5.The time required for the response curve to reach and stay within the specified 2-5% of final value is referred to as :
Q6.Consider a standard second order system given by \(\frac{w_{n}^{2}}{s^{2}+2\zeta w_{n} s+w_{n}^{2}}\). The correlation between the maximum peak overshoot in the time domain and the resonant peak in the frequency domain exists when:
Q7.Consider the second order closed loop control system. If the characteristics polynomial is given by S2 + 3S + 2 = 0, then the control system is:
Q8.A second order system has only imaginary conjugate poles, what is the damping factor value for this system?
Q9.For a second order underdamped control system, the settling time for the 5% band is given by the equation:
Q10.Consider a standard second order system given by \(\frac{w_{n}^{2}}{s^{2}+2 \zeta w_{n} s+w_{n}^{2}}\). The speed of response is measured in time and frequency domain by:

More Time Response Analysis Questions

Q1.A unity feedback system is characterized by the open-loop transfer function \(G(s)=\frac{100}{s(5 s+10)(2 s+10)}\) The steady-state errors for unit-step and unit-ramp inputs are :
Q2.In the circuit shown switch is closed at t = 0. The time constant of the circuit is :
Q3.Consider a causal second-order system with the transfer function \(G(s)=\frac{1}{s^2+2 s+1}\) with a unit-step \(R(s)=\frac{1}{s}\) as an input. Let c(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value \(\displaystyle \lim _{t \rightarrow \infty} c(t)\), rounded off to two decimal places, is
Q4.The output of a standard second-order system for a unit-step input is given as \(y(t)=1-\frac{2}{\sqrt{3}} e^{-t} \cos \left(\sqrt{3} t-\frac{\pi}{6}\right)\). What is the transfer function of the system?
Q5.A second-order system has a closed-loop transfer function given by \(G(s)=\frac{25}{s^2+8 s+25}\). The settling time for 5 percentage band in tolerance error is
Q6.A second-order system has a transfer function given by \(G(s)=\frac{25}{s^2+8 s+25}\). If the system, initially at rest, is subjected to a unit-step input at t = 0, the second peak in the response will occur at
Q7.Steady state error is the value of error signal when:
Q8.A closed loop system is shown in the figure where k > 0 and α > 0. The steady state error due to a ramp input (R(s) = α/s2) is given by 
Q9.The nature of the response shown below is:
Q10.When the damping ratio ζ (zeta) lies in between 0 < ζ < 1, then the circuit classification comes under:
Suggested Test Series
View All >
AE & JE Civil Engg. Exams Previous Year Papers Mock Test
469 Total Tests with 3 Free Tests
Start Free Test
AE & JE Electrical Engg. Exams Previous Year Papers Mock Test
276 Total Tests with   3 Free Tests
Start Free Test

More Control Systems Questions

Q1.A unity feedback system is characterized by the open-loop transfer function \(G(s)=\frac{100}{s(5 s+10)(2 s+10)}\) The steady-state errors for unit-step and unit-ramp inputs are :
Q2.In the circuit shown switch is closed at t = 0. The time constant of the circuit is :
Q3.The loop transfer function of a closed-loop control system is given as : \(G(s) H(s)=\frac{k(s+1)}{s(s+2)(s+3)}\) The centroid of the asymptotes in root locus is :
Q4.For a control system having fourteen poles and two zeros, the slope in its magnitude plot for high frequency asymptote will be :
Q5.The Routh Array is as below : \(\begin{array}{lllll} \mathrm{S}^{6} & 1 & 8 & 20 & 16 \\ \mathrm{S}^{5} & 2 & 12 & 16 & \\ \mathrm{S}^{4} & 2 & 12 & 16 & \\ \mathrm{S}^{3} & 0 & 0 & & \end{array}\) The row of zero of this array will be replaced by coefficients of 
Q6.The number of roots in the left half of the s-plane for a system having characteristic equations : s3 + 5s2 + 7s + 3 = 0 is : 
Q7.Explain the pattern and trend of public expenditure on social services in the post reforms period in India. To what extent this has been in consonance with achieving objective of inclusive growth? 
Q8.The Nyquist criterion for stability states that an amplifier is unstable if the Nyquist curve encloses the -1 + j0 point, and the amplifier is stable if the curve does not enclose this point. If Aβ
Q9.The phase margin of a system with the open-loop transfer function \(G(s) H(s)=\frac{1-s}{(s+1)(s+2)}\) is
Q10.The phase margin of a system having the loop transfer function \(G(s) H(s)=\frac{2 \sqrt{3}}{s(s+1)}\) is
Testbook Edu Solutions Pvt. Ltd.
1st & 2nd Floor, Zion Building,
Plot No. 273, Sector 10, Kharghar,
Navi Mumbai - 410210
support@testbook.com
Toll Free:1800 833 0800
Office Hours: 10 AM to 7 PM (all 7 days)
Company
About usCareers We are hiringTeach Online on TestbookPartnersMediaSitemap
Products
Test SeriesTestbook PassOnline CoursesOnline VideosPracticeBlogRefer & EarnBooks
Copyright © 2014-2022 Testbook Edu Solutions Pvt. Ltd.: All rights reserved
User PolicyTermsPrivacy
Hot Links： teen patti master purana teen patti comfun card online teen patti cash