Forced Vibration and Resonance Questions and Answers

Answer: C

The match is (P) → (i), (Q) → (iv), (R) → (ii), (S) → (iii).

• (P) \(c \gt 0\), \(\omega = \sqrt{k/m}\): damped resonance. The amplitude grows and then settles at a finite steady value, \(B/(c\omega)\), because damping removes energy as fast as the force supplies it. That is response (i), whose envelope levels off.
• (Q) \(c \lt 0\): negative damping feeds energy in, so the free response grows exponentially whatever the forcing frequency. That is (iv), whose envelope curves outwards.
• (R) \(c = 0\), \(\omega = \sqrt{k/m}\): undamped resonance. The particular solution is \(x \propto t\cos\omega_{n}t\), so the amplitude grows linearly without limit. That is (ii), with straight-line envelopes.
• (S) \(c = 0\), \(\omega \cong \sqrt{k/m}\): forcing close to, but not at, the natural frequency. The free and forced vibrations drift in and out of phase, producing beats. That is (iii).

The other options swap these pairs; the quickest check is that the beat pattern (iii) can only belong to the near-resonance case (S), which leaves only one option.

Answer: A

The decaying part of the response is the transient response.

Under harmonic forcing, the motion is the sum of two parts. The complementary solution is the system's free vibration, which carries a factor \(e^{-\zeta\omega_{n}t}\) and so dies away with time because of damping: this is the transient. The particular solution oscillates at the forcing frequency with constant amplitude for as long as the force acts: this is the steady-state response.

So the steady-state (non-transient) part is precisely the part that does not decay, and 'harmonic response' describes the forced oscillation at the driving frequency, again the part that persists.

Answer: D

Resonance occurs at \(\omega = \sqrt{K/M}\).

Resonance is excitation at the natural frequency of the undamped system. The free equation \(M\ddot{x} + Kx = 0\) gives \(\omega_{n} = \sqrt{K/M}\) in rad/s, and the forced amplitude \(F/(K - M\omega^{2})\) grows without limit as \(\omega\) approaches it.

\(\frac{1}{2\pi}\sqrt{K/M}\) is the same natural frequency in hertz, but \(\omega\) is an angular frequency, so the units do not match. \(2\pi\sqrt{K/M}\) multiplies where it should not, and \(\sqrt{M/K}\) is inverted (it has units of time).

Answer: A

The correct order is \(\omega_{p} \lt \omega_{d} \lt \omega_{n}\).

Minimising \((1 - r^{2})^{2} + (2\zeta r)^{2}\) with respect to \(r^{2}\) gives the peak of the amplitude curve:

\[ \begin{aligned} r_{p}^{2} &= 1 - 2\zeta^{2} \Rightarrow \omega_{p} = \omega_{n}\sqrt{1 - 2\zeta^{2}} \\ \omega_{d} &= \omega_{n}\sqrt{1 - \zeta^{2}} \end{aligned} \]

For any damping \(0 \lt \zeta \lt 1/\sqrt{2}\), \(1 - 2\zeta^{2} \lt 1 - \zeta^{2} \lt 1\), so the peak lies below the damped natural frequency, which lies below the undamped one.

\(\omega_{p} = \omega_{d}\) is a common confusion: the free vibration oscillates at \(\omega_{d}\), but the forced amplitude peaks lower. The peak equals \(\omega_{n}\) only with no damping and never exceeds it, which rules out the last two options.