# Forced Vibration

1. Jun 24, 2005

### 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?

2. Jun 25, 2005

### OlderDan

(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

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

3. Jun 25, 2005

### e(ho0n3

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

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