Undamped System MCQ Quiz - Objective Question with Answer for Undamped System - Download Free PDF

Concept:

Time-domain specification (or) transient response parameters:

Rise time (t­r): It is the time taken by the response to reach from 0% to 100% Generally 10% to 9% for overdamped and 5% to 95% for the critically damped system is defined.

$c\left(t\right){|}_{t=t_r}=1-\frac{e^{-\xi {\omega }_n{t}_r}}{\sqrt{1-{\xi }^2}}\sin \left({\omega }_n{t}_r+\phi \right)$

$t_r=\frac{\pi -\phi }{{\omega }_d}$

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

${\frac{dc\left(t\right)}{dt}|}_{t=t_p}=0,t_p=\frac{\pi }{{\omega }_d}$

Calculation:

Rise time $\left(t_r\right)=\frac{\pi -\varphi }{{\omega }_d}$

Peat time $\left(t_p\right)=\frac{\pi }{{\omega }_d}$

t_p = 2 t_r

$\Rightarrow \frac{\pi }{{\omega }_d}=\frac{2\left(\pi -\varphi \right)}{{\omega }_d}$

$\Rightarrow \frac{\pi }{2}=\pi -\varphi \Rightarrow \varphi =\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}\left(\frac{\sqrt{1-{\xi }^2}}{\xi }\right)=\frac{\pi }{2}$

$\Rightarrow \frac{\sqrt{1-{\xi }^2}}{\xi }=\mathrm{\infty }$

⇒ ξ = 0

Therefore, the system is undamped.

Delay time (td): It is the time taken by the response to change from 0 to 50% of its final or steady-state value.

$c\left(t\right){|}_{t=t_d}=0.5$

$t_d\simeq \frac{1+0.7\xi }{{\omega }_n}$

Maximum (or) Peak overshoot (Mp): It is the maximum error at the output. 

$M_p=c\left(t_p\right)-1,M_p=e^{\left(\frac{-\xi \pi }{\sqrt{1-{\xi }^2}}\right)}$

$\mathrm{\%}{M}_p=\frac{c\left(t_p\right)-c\left(\mathrm{\infty }\right)}{c\left(\mathrm{\infty }\right)}\times 100\mathrm{\%}$

If the magnitude of the input is doubled, then the steady-state value doubles, therefore Mp doubles, but % Mp, tr, tp remains constant.

Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.

$e^{-\xi {\omega }_n{t}_s}=\pm 5\mathrm{\%}\left(or\right)\pm 2\mathrm{\%}$

 $t_s\simeq \frac{3}{\xi {\omega }_n}$ for a 5% tolerance band.

$t_s\simeq \frac{4}{\xi {\omega }_n}$ for 2% tolerance band

If ξ increases, rise time, peak time increases and peak overshoot decreases.

The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force.

Since for the given waveform, the amplitude of the natural oscillation is decreasing and increasing, the oscillations are not damped, because                    damped oscillations gradually decay to zero with time as shown:

Beats occur is a mechanical system when we excite it with vibrations that are close to its natural frequency.

Taking an example of an undammed, forced oscillation mass-spring system as shown:

When the input frequency (excitation frequency) ω is close to, but not quite equal to the natural frequency ω₀, the mass m will move up and down is the manner as shown:

The frequency $\frac{{\omega }_0-\omega }{2}$ is called the envelope frequency.

From this, we can conclude that the given oscillations are that of an undamped forced vibration.

We know that characteristic equation.

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

$1+\frac{k\left(1+k_{P}s\right)}{s\left(s+1\right)}=0$

s² + s (1 + k_Kp) + k = 0

s² + 2ζω_ns + ω²n = 0

${\omega }_n=\sqrt{k}=4$

k = 16

2ζωn = 1+ kK_p

1 + kK_p = 2 × 0.8 × 4 = 6.4

kK_p = 5.4

The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force.

Since for the given waveform, the amplitude of the natural oscillation is decreasing and increasing, the oscillations are not damped, because                    damped oscillations gradually decay to zero with time as shown:

Beats occur is a mechanical system when we excite it with vibrations that are close to its natural frequency.

Taking an example of an undammed, forced oscillation mass-spring system as shown:

When the input frequency (excitation frequency) ω is close to, but not quite equal to the natural frequency ω₀, the mass m will move up and down is the manner as shown:

The frequency $\frac{{\omega }_0-\omega }{2}$ is called the envelope frequency.

From this, we can conclude that the given oscillations are that of an undamped forced vibration.

We know that characteristic equation.

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

$1+\frac{k\left(1+k_{P}s\right)}{s\left(s+1\right)}=0$

s² + s (1 + k_Kp) + k = 0

s² + 2ζω_ns + ω²n = 0

${\omega }_n=\sqrt{k}=4$

k = 16

2ζωn = 1+ kK_p

1 + kK_p = 2 × 0.8 × 4 = 6.4

kK_p = 5.4

Concept:

Time-domain specification (or) transient response parameters:

Rise time (t­r): It is the time taken by the response to reach from 0% to 100% Generally 10% to 9% for overdamped and 5% to 95% for the critically damped system is defined.

$c\left(t\right){|}_{t=t_r}=1-\frac{e^{-\xi {\omega }_n{t}_r}}{\sqrt{1-{\xi }^2}}\sin \left({\omega }_n{t}_r+\phi \right)$

$t_r=\frac{\pi -\phi }{{\omega }_d}$

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

${\frac{dc\left(t\right)}{dt}|}_{t=t_p}=0,t_p=\frac{\pi }{{\omega }_d}$

Calculation:

Rise time $\left(t_r\right)=\frac{\pi -\varphi }{{\omega }_d}$

Peat time $\left(t_p\right)=\frac{\pi }{{\omega }_d}$

t_p = 2 t_r

$\Rightarrow \frac{\pi }{{\omega }_d}=\frac{2\left(\pi -\varphi \right)}{{\omega }_d}$

$\Rightarrow \frac{\pi }{2}=\pi -\varphi \Rightarrow \varphi =\frac{\pi }{2}$

$\Rightarrow {\tan }^{-1}\left(\frac{\sqrt{1-{\xi }^2}}{\xi }\right)=\frac{\pi }{2}$

$\Rightarrow \frac{\sqrt{1-{\xi }^2}}{\xi }=\mathrm{\infty }$

⇒ ξ = 0

Therefore, the system is undamped.

Delay time (td): It is the time taken by the response to change from 0 to 50% of its final or steady-state value.

$c\left(t\right){|}_{t=t_d}=0.5$

$t_d\simeq \frac{1+0.7\xi }{{\omega }_n}$

Maximum (or) Peak overshoot (Mp): It is the maximum error at the output. 

$M_p=c\left(t_p\right)-1,M_p=e^{\left(\frac{-\xi \pi }{\sqrt{1-{\xi }^2}}\right)}$

$\mathrm{\%}{M}_p=\frac{c\left(t_p\right)-c\left(\mathrm{\infty }\right)}{c\left(\mathrm{\infty }\right)}\times 100\mathrm{\%}$

If the magnitude of the input is doubled, then the steady-state value doubles, therefore Mp doubles, but % Mp, tr, tp remains constant.

Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.

$e^{-\xi {\omega }_n{t}_s}=\pm 5\mathrm{\%}\left(or\right)\pm 2\mathrm{\%}$

 $t_s\simeq \frac{3}{\xi {\omega }_n}$ for a 5% tolerance band.

$t_s\simeq \frac{4}{\xi {\omega }_n}$ for 2% tolerance band

If ξ increases, rise time, peak time increases and peak overshoot decreases.

The natural frequency is the frequency at which a system would oscillate if there were no driving and no damping force.

Since for the given waveform, the amplitude of the natural oscillation is decreasing and increasing, the oscillations are not damped, because                    damped oscillations gradually decay to zero with time as shown:

Beats occur is a mechanical system when we excite it with vibrations that are close to its natural frequency.

Taking an example of an undammed, forced oscillation mass-spring system as shown:

When the input frequency (excitation frequency) ω is close to, but not quite equal to the natural frequency ω₀, the mass m will move up and down is the manner as shown:

The frequency $\frac{{\omega }_0-\omega }{2}$ is called the envelope frequency.

From this, we can conclude that the given oscillations are that of an undamped forced vibration.

We know that characteristic equation.

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

$1+\frac{k\left(1+k_{P}s\right)}{s\left(s+1\right)}=0$

s² + s (1 + k_Kp) + k = 0

s² + 2ζω_ns + ω²n = 0

${\omega }_n=\sqrt{k}=4$

k = 16

2ζωn = 1+ kK_p

1 + kK_p = 2 × 0.8 × 4 = 6.4

kK_p = 5.4

$\begin{array}{l}{\omega }_n=\sqrt{10}rad/sec\\ 2\zeta {\omega }_n=4\\ \Rightarrow \zeta =\frac{2}{\sqrt{10}}\end{array}$

Damped natural frequency, ${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}=\sqrt{6}$

$\Rightarrow f_d=\frac{\sqrt{6}}{2\pi }Hz$

CE is given by $1+\frac{\mathrm{K}\left(1-\mathrm{s}\right)}{\mathrm{s}\left(\mathrm{s}+2\right)}=0$

${\mathrm{s}}^2+\left(2-\mathrm{K}\right)\mathrm{s}+\mathrm{K}=0$

R-H criterion →

$\begin{array}{c}{s}^2\\ s^1\\ s^2\end{array}|\begin{array}{rr}1& \mathrm{K}\\ 2-\mathrm{K}& \\ \mathrm{K}& \end{array}$

for the intersection with imaginary axis, we should have $2-\mathrm{K}=0\to \mathrm{K}=2$

auxiliary equation $\to {\mathrm{s}}^2+\mathrm{K}=0$

$\begin{array}{l}{s}^2+2=0\\ \to s=j\pm \sqrt{2}\end{array}$