Question

A popular undergrad classical mechanics text (JR Taylor, 2005) expresses the following eqn. for the DDHO: ï + 2Bi +w²x = f. cos(wt). Further, they express a general (steady-state) solution as: x(t) = A cos(wt – 8) a) Show that the amplitude (squared) is given by (62 – w2)2 + 48202 b) Verify that the following is in fact true: W2 = V02 – 282 = value of w at which response is maximum.

Answer 1

x + 2 Bu +wor - to Cosut – Re [to e Jwt ] in Steady steady sto state the system Vibrates force, we may try since frequency of driving a solution of the form x = Re [xo e swt ] substituting we get, Re[w? 26 w j + wir xo ejut – Reto lint This is is value of t and hence Yo true for all to (W, 2w 2) tj 2 BW - Ae-ja amplitude A (wr wr) + 4own A2 (wewn)2 + 4Bw? The amplitude A will be maximum when dA 0 on 2(W6-w) (-2W) +88 or wh- w - 282 W= I wa_282

Amplitude Amay when we wo%2B2 is (wor ws +262) 2 + 48* (w. - 282) ✓ Aprw 2ßwo associated Resonant & value frequency band width wo 28 @ Time average of power (87= mBrwn A2 = nB wt to? ( Wyrwyr ABW2 <py = max when wzwo. <P> max = mB for 482 . L87 = {l>max 40 we (Wown) 2 At half power point LP>= LP) mex 2 or worn wr= 126 or w. -8 + J 8² + us and we = Wow, = 2B p + JBn won