Resonant Frequency MCQ Quiz - Objective Question with Answer for Resonant Frequency - Download Free PDF

Last updated on Apr 9, 2025

Latest Resonant Frequency MCQ Objective Questions

Resonant Frequency Question 1:

A second-order system has

C(s)R(s)=ωn2(s2+2ξωns+ωn2)

Its frequency response will have a maximum value at the frequency:

  1. ωn(1ξ2)
  2. ωnξ
  3. ωn(12ξ2)
  4. Zero

Answer (Detailed Solution Below)

Option 3 : ωn(12ξ2)

Resonant Frequency Question 1 Detailed Solution

The transfer function of the standard 2nd order system is defined as

T.F=ωn2s2+2ξωns+ωn2

The resonant frequency ω0 is given by

ω0=ωn12ξ2

Where ωn = undamped natural frequency

ξ = Damping ratio

Resonant peak M0 is given by:

M0=12ξ1ξ2,0ξ12

Resonant frequency (ωr) is the frequency at which the magnitude characteristic of frequency response curve has its peak value.

The above equation is meaningful only for 1 – 2ζ2 ≥ 0, i.e.

zeta12 or ζ0.707

Note: If ζ > 0.707, then ωr = 0

Resonant Frequency Question 2:

Match the following

1) Resonant frequency        P)  ωn12ξ2+24ξ2+4ξ4
2) Bandwidth       Q)  ωn12ξ2
3) Resonant peak    R)  ωn1ξ2
4) Damping ratio S) 12ξ1ξ2
  T) 12Q

   

                                                          

 

  1. 1)Q,2)P,3)S,4)T

  2. 1)R,2)P,3)S,4)T

  3. 1)Q,2)P,3)R,4)T

  4. 1)R,2)P,3)Q,4)T

Answer (Detailed Solution Below)

Option 1 :

1)Q,2)P,3)S,4)T

Resonant Frequency Question 2 Detailed Solution

Resonant frequency, ωr=ωn12ξ2

Damping Ratio ξ=12Q

Resonant Peak  Mr=12ξ1ξ2

Bandwidth BW=ωn12ξ2+24ξ2+4ξ4

Top Resonant Frequency MCQ Objective Questions

A second-order system has

C(s)R(s)=ωn2(s2+2ξωns+ωn2)

Its frequency response will have a maximum value at the frequency:

  1. ωn(1ξ2)
  2. ωnξ
  3. ωn(12ξ2)
  4. Zero

Answer (Detailed Solution Below)

Option 3 : ωn(12ξ2)

Resonant Frequency Question 3 Detailed Solution

Download Solution PDF

The transfer function of the standard 2nd order system is defined as

T.F=ωn2s2+2ξωns+ωn2

The resonant frequency ω0 is given by

ω0=ωn12ξ2

Where ωn = undamped natural frequency

ξ = Damping ratio

Resonant peak M0 is given by:

M0=12ξ1ξ2,0ξ12

Resonant frequency (ωr) is the frequency at which the magnitude characteristic of frequency response curve has its peak value.

The above equation is meaningful only for 1 – 2ζ2 ≥ 0, i.e.

zeta12 or ζ0.707

Note: If ζ > 0.707, then ωr = 0

Resonant Frequency Question 4:

A second-order system has

C(s)R(s)=ωn2(s2+2ξωns+ωn2)

Its frequency response will have a maximum value at the frequency:

  1. ωn(1ξ2)
  2. ωnξ
  3. ωn(12ξ2)
  4. Zero

Answer (Detailed Solution Below)

Option 3 : ωn(12ξ2)

Resonant Frequency Question 4 Detailed Solution

The transfer function of the standard 2nd order system is defined as

T.F=ωn2s2+2ξωns+ωn2

The resonant frequency ω0 is given by

ω0=ωn12ξ2

Where ωn = undamped natural frequency

ξ = Damping ratio

Resonant peak M0 is given by:

M0=12ξ1ξ2,0ξ12

Resonant frequency (ωr) is the frequency at which the magnitude characteristic of frequency response curve has its peak value.

The above equation is meaningful only for 1 – 2ζ2 ≥ 0, i.e.

zeta12 or ζ0.707

Note: If ζ > 0.707, then ωr = 0

Resonant Frequency Question 5:

A unit step input is applied to a unity feedback control system having open-loop transfer function,

G(s)H(s)=Ks(1+sT)

The specification is such that the maximum overshoot is 20% and the resonant frequency is ωr = 6 rad/sec. The values of K and T are

  1. 6.71, 1.4
  2. 8.67, 0.14
  3. 0.14, 8.67
  4. 1.4, 6.71

Answer (Detailed Solution Below)

Option 2 : 8.67, 0.14

Resonant Frequency Question 5 Detailed Solution

The characteristic equation for the given open-loop transfer function is defined as:

1 + G(s)H(s) = 0

∴ For the given open-loop transfer function, we will get the characteristic equation as:

s2+1Ts+KT=0     ---(1)

The standard form of the characteristic equation of the second-order system will be:

s2+2ξωns+ωn2=0

Comparing this with Equation (1), we get:

ωn=KT    ---(2)

Also 2ξωn=1T

Using Equation (2), we get:

ξ=12KT      ---(3)

From the given problem, we have the maximum overshoot as:

Mp = 20%

Since the maximum overshoot is defined as:

Mp=ξπe1ξ2=20100

Putting on the respective values:

ξπ1ξ2=1.6

ξ = 0.45

Now, the resonant frequency is given by:

ωr=ωn12ξ2=6

Therefore, the natural frequency will be:

ωn=612(0.45)2

= 7.82 rad/sec

Substituting the values of ξ and ωn in Equations (1) and (2), we get:

(7.82)2=KTand

0.45=12KT

Solving the above equations, we obtain the values of K and T as:

K = 8.67 and T = 0.14 

Resonant Frequency Question 6:

Match the following

1) Resonant frequency        P)  ωn12ξ2+24ξ2+4ξ4
2) Bandwidth       Q)  ωn12ξ2
3) Resonant peak    R)  ωn1ξ2
4) Damping ratio S) 12ξ1ξ2
  T) 12Q

   

                                                          

 

  1. 1)Q,2)P,3)S,4)T

  2. 1)R,2)P,3)S,4)T

  3. 1)Q,2)P,3)R,4)T

  4. 1)R,2)P,3)Q,4)T

Answer (Detailed Solution Below)

Option 1 :

1)Q,2)P,3)S,4)T

Resonant Frequency Question 6 Detailed Solution

Resonant frequency, ωr=ωn12ξ2

Damping Ratio ξ=12Q

Resonant Peak  Mr=12ξ1ξ2

Bandwidth BW=ωn12ξ2+24ξ2+4ξ4

Resonant Frequency Question 7:

Match the following

1)         Resonant frequency               P)      ωn12ξ2+24ξ2+4ξ4

2)         Bandwidth                              Q)      ωn12ξ2

3)         Resonant peak                      R)      ωn1ξ2

4)         Damping ratio                        S)      12ξ1ξ2

                                                          T)      12Q

 

  1. 1)Q,2)P,3)S,4)T
  2. 1)R,2)P,3)S,4)T
  3. 1)Q,2)P,3)R,4)T
  4. 1)R,2)P,3)Q,4)T
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 1)Q,2)P,3)S,4)T

Resonant Frequency Question 7 Detailed Solution

Resonant frequency, ωr=ωn12ξ2

Damping Ratio ξ=12Q

Resonant Peak  Mr=12ξ1ξ2

Bandwidth BW=ωn12ξ2+24ξ2+4ξ4

Resonant Frequency Question 8:

Range of ϕr with ξ variation is –

r = phase at resonant freq.)

  1. -90° to 0° 
  2. +90° to 0° 
  3. -180° to 0°       
  4. +180° to 0°
  5. -90° to 90° 

Answer (Detailed Solution Below)

Option 1 : -90° to 0° 

Resonant Frequency Question 8 Detailed Solution

Range ξ is 0<ξ<1/2

ϕr|ξ=0=tan1[12ξ2ξ]=90

ϕr|ξ=12=tan1[12ξ2ξ]=0

Resonant Frequency Question 9:

Range of ϕr with ξ variation is –

r = phase at resonant freq.)

  1. -90° to 0° 
  2. +90° to 0° 
  3. -180° to 0°       
  4. +180° to 0°

Answer (Detailed Solution Below)

Option 1 : -90° to 0° 

Resonant Frequency Question 9 Detailed Solution

Range ξ is 0<ξ<1/2

ϕr|ξ=0=tan1[12ξ2ξ]=90

ϕr|ξ=12=tan1[12ξ2ξ]=0

Resonant Frequency Question 10:

Phase at resonant frequency if ξ = 0.4 is 

  1. 66.42° 
  2. 64.12° 
  3. -66.42° 
  4. -64.12° 

Answer (Detailed Solution Below)

Option 4 : -64.12° 

Resonant Frequency Question 10 Detailed Solution

ϕ=tan1[12ξ2ξ]

=64.12°

since control systems are Low pass filters that give more negative phase. So Ø = -64.12 degrees

Note :

Cos-1ξ  cannot be used here because this is resonance condition

Resonant Frequency Question 11:

A second-order system has

C(s)R(s)=ωn2(s2+2ξωns+ωn2)

Its frequency response will have a maximum value at the frequency:

  1. ωn(1ξ2)
  2. ωnξ
  3. ωn(12ξ2)
  4. Zero
  5. None of these

Answer (Detailed Solution Below)

Option 3 : ωn(12ξ2)

Resonant Frequency Question 11 Detailed Solution

The transfer function of the standard 2nd order system is defined as

T.F=ωn2s2+2ξωns+ωn2

The resonant frequency ω0 is given by

ω0=ωn12ξ2

Where ωn = undamped natural frequency

ξ = Damping ratio

Resonant peak M0 is given by:

M0=12ξ1ξ2,0ξ12

Resonant frequency (ωr) is the frequency at which the magnitude characteristic of frequency response curve has its peak value.

The above equation is meaningful only for 1 – 2ζ2 ≥ 0, i.e.

zeta12 or ζ0.707

Note: If ζ > 0.707, then ωr = 0

Resonant Frequency Question 12:

An under-damped second order system having a transfer function of the form:

T(s)=10Ks2+2s+K

has a frequency response plot as shown below:

F3 S.B 21.8.20 Pallavi D2

Which of the following statements is/are true?

  1. The resonant peak is 20
  2. The value of K is 10
  3. The natural frequency is 3.8 rad/s
  4. The damping factor ζ is 0.965

Answer (Detailed Solution Below)

Option :

Resonant Frequency Question 12 Detailed Solution

Concept:

DC gain: DC Gain of a system is the gain at the steady-state which is at t tending to infinity i.e., s tending to zero.

Resonant peak (Mr): The maximum value of the magnitude of closed-loop transfer function in the frequency response is called the resonant peak.

Mr=DCgain2ζ1ζ2

Resonant frequency (ωr): The frequency at which the resonant peak occurs is termed as the resonant frequency.

ωr=ωn12ζ2

Calculation:

DC gain calculation:

T(0)=10KK=10

Resonant peak = 20 (as shown in the figure)

Damping factor calculation:

20=102ζ1ζ2

4ζ1ζ2=1

16ζ2(1ζ2)=1

Let ζ2 = x

16x (1-x) = 1

16x2 – 16x + 1 = 0

x=16±1624×162×16

X = 0.933 and x = 0.067

ζ = 0.965 and ζ = 0.259

If resonant peak > 1 then ζ12=0.707

ζ = 0.259

Gain and natural frequency calculation:

The standard characteristic equation of a second order system is:

s2 + 2ζ ωn s + ωn2 = 0

Comparing the characteristic equation of the given T.F:

s2 + 2ζ ωn s + ωn2 = s2 + 2s + K

ωn = √ K

2 ζ ωn = 2

2 ζ √ K = 2

K = 1ζ2=10.067=15.02

ωn = √ 15.02 = 3.8 rad/s

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