french5 - 1 University of California, Berkeley Physics H7A...

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1 University of California, Berkeley Physics H7A Fall 1998 ( Strovink ) SOLUTION TO PROBLEM SET 10 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. French 5-6. (a.) All three springs are identical, with constant k . The equations of motion are d 2 x A dt 2 +2 ω 2 0 x A ω 2 0 x B =0 d 2 x B dt 2 +2 ω 2 0 x B ω 2 0 x A =0 We plug in a guessed solution, where the two masses oscillate at the same frequency, but with di±erent amplitudes A and B . This gives ω 2 A +2 ω 2 0 A ω 2 0 B =0 B = A 2 ω 2 0 ω 2 ω 2 0 ω 2 B +2 ω 2 0 B ω 2 0 A =0 B = A ω 2 0 2 ω 2 0 ω 2 Equating these we see that (2 ω 2 0 ω 2 ) 2 ω 4 0 =0 Solving this quadratic equation, we ²nd that the two frequencies are ω 2 = ω 2 0 and ω 2 =3 ω 2 0 . (b.) One mass is displaced by 5 cm. This excites each normal mode equally, with amplitude 2.5 cm. To see this, ²rst excite the ²rst normal mode with amplitude 2.5 cm. Now both masses are +2.5 cm from equilibrium. Now excite the sec- ond normal mode, also with amplitude 2.5 cm. This moves one mass forward 2.5 cm, and the other back 2.5 cm. One is now 5cm from equilib- rium, and the other is at its equilibrium position. It is mass B that is displaced, so the masses obey x A =2 . 5cos ω 0 t 2 . 5cos 3 ω 0 t x B =2 . 5cos ω 0 t +2 . 5cos 3 ω 0 t (c.) After a time τ such that cos ω 0 τ = cos 3 ω 0 τ , mass A returns to its equilibrium position x A = 0. This happens when ω 0 τ is in the second quadrant and 3 ω 0 τ is in the third: π ω 0 τ = 3 ω 0 τ π ω 0 τ = 2 π 1+ 3 However, at t = τ , mass B will not have re- turned to its full original | displacement | since | cos
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french5 - 1 University of California, Berkeley Physics H7A...

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