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GATE 2022 ME (ME1) – Question 39

Vibrations · Forced vibration and resonance · 2 marks · Multiple choice

Consider a forced single degree-of-freedom system governed by $\ddot{x}(t) + 2\zeta\omega_n\dot{x}(t) + \omega_n^2x(t) = \omega_n^2\cos(\omega t)$, where $\zeta$ and $\omega_n$ are the damping ratio and undamped natural frequency of the system, respectively, while $\omega$ is the forcing frequency. The amplitude of the forced steady state response of this system is given by $[(1 - r^2)^2 + (2\zeta r)^2]^{-1/2}$, where $r = \omega/\omega_n$. The peak amplitude of this response occurs at a frequency $\omega = \omega_p$. If $\omega_d$ denotes the damped natural frequency of this system, which one of the following options is true?

  1. $\omega_p < \omega_d < \omega_n$
  2. $\omega_p = \omega_d < \omega_n$
  3. $\omega_d < \omega_n = \omega_p$
  4. $\omega_d < \omega_n < \omega_p$

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Correct answer: (A) $\omega_p < \omega_d < \omega_n$

Explanation

The amplitude is maximum when $\frac{d}{dr}\left[(1 - r^2)^2 + 4\zeta^2r^2\right] = 0$, which gives $r^2 = 1 - 2\zeta^2$, so $\omega_p = \omega_n\sqrt{1 - 2\zeta^2}$. The damped natural frequency is $\omega_d = \omega_n\sqrt{1 - \zeta^2}$. For $\zeta > 0$: $\sqrt{1 - 2\zeta^2} < \sqrt{1 - \zeta^2} < 1$, so $\omega_p < \omega_d < \omega_n$.