Question

A system is described by the following differential equation: Ï + + 4 x = sin(at) The steady-state response can be written: Xss = A sinot + 0) If the phase angle is: JT O = radians 2 What is the excitation frequency, o ? rad 0 = 0 S rad w = 1 S rad 0 = 2 S rad 0 = 4

Answer 1

Solution:-

  $\ddot{x}+\dot{x}+4x=0............................Equation \ (1)$

  $\ddot{x}+2\xi \omega _{n}\dot{x}+\omega _{n}^{2}x=0............................Equation \ (2)$

Compare equation (1) with equation (2)

  $\omega _{n}^{2}=4$

after solving   

$\omega _{n}=2 \ \ rad/sec$

now compare

  $2\xi \omega _{n}=1$

  $2\xi \times 2=1$

  $\xi =\frac{1}{4}$

Now calcualte excitation frequency

$\omega _{d}=\omega _{n}\sqrt{1-\xi ^{2}}$

$\omega _{d}=2\sqrt{1-\left ( \frac{1}{4} \right ) ^{2}}$​​​​​​​

$\omega _{d}=1.936 \ rad/sec$

Approximate answer is option c (2 rad/sec)...................................Ans

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