Forced Vibration Questions: Solve Equations for Answers

[e(ho0n3]

Questions:
(a) At what resonance (\omega = \omega_0), what is the value of the phase angle \phi?
(b) What, then, is the displacement at a time when the driving force F_{ext} is a maximum, and at a time when F_{ext} = 0?
(c) What is the phase difference (in degrees) between the driving force and the displacement in this case?

Equations related to this problem:

F_{ext} = F_0\cos{\omega t}

x = A_0\sin{(\omega t + \phi_0)}

\phi_0 = \tan^{-1}\frac{\omega_0^2 - \omega^2}{\omega(b/m)}

My Answers:
(a) Since \omega = \omega_0, \phi_0 = \tan^{-1}0 which means \phi_0 = k\pi for some non-negative integer k.
(b) F_{ext} has its maximum value when \omega t = 2j\pi for some non-negative integer j. The displacement is then x = A_0\sin{(2j\pi + k\pi)} = 0. F_{ext} = 0 implies that \omega t = l\pi/2 where l is some odd positive integer.The displacement is then x = A_0\sin{(l\pi/2 + k\pi)}, so x = A₀ or -A₀.
(c) This question I don't understand well. I'm guess the difference is \pi/2 + \phi_0 because the driving force is a cosine function and the displacement is a sine function with a phase angle.

Is this right?

 

[OlderDan]

Questions:
(a) At what resonance (\omega = \omega_0), what is the value of the phase angle \phi?
(b) What, then, is the displacement at a time when the driving force F_{ext} is a maximum, and at a time when F_{ext} = 0?
(c) What is the phase difference (in degrees) between the driving force and the displacement in this case?

Equations related to this problem:

F_{ext} = F_0\cos{\omega t}

x = A_0\sin{(\omega t + \phi_0)}

\phi_0 = \tan^{-1}\frac{\omega_0^2 - \omega^2}{\omega(b/m)}

My Answers:
(a) Since \omega = \omega_0, \phi_0 = \tan^{-1}0 which means \phi_0 = k\pi for some non-negative integer k.
(b) F_{ext} has its maximum value when \omega t = 2j\pi for some non-negative integer j. The displacement is then x = A_0\sin{(2j\pi + k\pi)} = 0. F_{ext} = 0 implies that \omega t = l\pi/2 where l is some odd positive integer.The displacement is then x = A_0\sin{(l\pi/2 + k\pi)}, so x = A₀ or -A₀.
(c) This question I don't understand well. I'm guess the difference is \pi/2 + \phi_0 because the driving force is a cosine function and the displacement is a sine function with a phase angle.

Is this right?

(c) What is the phase difference (in degrees) between the driving force and the displacement in this case?

Assuming your x equation is correct, and using

\sin{\alpha} = \cos{(\alpha - 90)}

you have

x = A_0\sin{(\omega t + \phi_0)} = A_0\cos{(\omega t + \phi_0 - 90)}

 

[e(ho0n3]

Ah, so the phase difference is just \phi_0 - \pi/2 radians. Thanks.