Question

Consider a single degree of freedom system with viscous damping excited by a harmonic force. At resonance, the phase angle of the displacement with respect to the exciting force in degrees is

• A. 0
• B. 45
• C. 90
• D. 135

Hint:

Phase angle (\(\phi\)) transitions relative to frequency ratio (\(r = \frac{\omega}{\omega_n}\)): - Low frequencies (\(r \ll 1\)): \(\phi \to 0^\circ\) (Response is in phase with force). - At Resonance (\(r = 1\)): \(\phi = 90^\circ\) exactly, regardless of the value of damping ratio \(\zeta\) (provided \(\zeta \gt 0\)). - High frequencies (\(r \gg 1\)): \(\phi \to 180^\circ\) (Response is out of phase).

Answer 1

The Correct Option is C

Concept: Consider a forced, single-degree-of-freedom mechanical system under viscous damping governed by the classical second-order ordinary differential equation: \[ m\ddot{x} + c\dot{x} + kx = F_0 \sin(\omega t) \] Where:

• \(m\) = Mass of the system

• \(c\) = Viscous damping coefficient

• \(k\) = Stiffness of the supporting spring element

• \(F_0\) = Peak amplitude of the external harmonic excitation force

• \(\omega\) = Operating frequency of the exciting force
The steady-state harmonic displacement response lags behind the driving force by a specific phase angle \(\phi\), written as: \[ x(t) = X \sin(\omega t - \phi) \] The phase angle \(\phi\) is expressed analytically as a function of the excitation frequency \(\omega\) and system parameters: \[ \tan\phi = \frac{c\omega}{k - m\omega^2} \] To generalize this behavior for any system, we divide the numerator and denominator by the stiffness \(k\): \[ \tan\phi = \frac{\frac{c\omega}{k}}{1 - \frac{m\omega^2}{k}} \] We define the natural undamped frequency as \(\omega_n = \sqrt{\frac{k}{m}}\) (so \(\frac{m}{k} = \frac{1}{\omega_n^2}\)) and the critical damping coefficient as \(c_c = 2\sqrt{km} = 2m\omega_n\). Introducing the non-dimensional parameters:

• Frequency ratio: \(r = \frac{\omega}{\omega_n}\)

• Damping ratio: \(\zeta = \frac{c}{c_c} = \frac{c}{2m\omega_n}\)
Substituting these terms provides the canonical phase angle equation: \[ \tan\phi = \frac{2\zeta \left(\frac{\omega}{\omega_n}\right)}{1 - \left(\frac{\omega}{\omega_n}\right)^2} = \frac{2\zeta r}{1 - r^2} \]

Step 1: Applying the condition of resonance.
Resonance is defined as the state where the frequency of the external harmonic force matches the natural undamped frequency of the mechanical system. Therefore: \[ \omega = \omega_n \quad \Rightarrow \quad r = \frac{\omega}{\omega_n} = 1 \]

Step 2: Computing the value of \(\tan\phi\) at \(r = 1\).
Substituting \(r = 1\) into our non-dimensional equation yields: \[ \tan\phi = \frac{2\zeta (1)}{1 - (1)^2} = \frac{2\zeta}{0} \to \infty \]

Step 3: Solving for the phase angle \(\phi\).
We find the inverse tangent of infinity: \[ \phi = \arctan(\infty) = \frac{\pi}{2} \text{ radians} = 90^\circ \] This mathematically demonstrates that at resonance, the displacement response delays behind the excitation force vector by exactly a quarter cycle (\(90^\circ\)), meaning the exciting force works entirely to overcome the viscous damping force. This matches Option (3).