**Unformatted text preview: **frequency is again Hz. 800 . The sack falls off when the spring is stretched 0.6895 m. And the speed of the beam at this point is ( 29 ( 29 ( 29 m/s. 508 . 1 kg 400 N/m 5685 m 400 . = = = = m k A v . Take = y at this point. The total energy of the beam at this point, just after the sack falls off, is ( 29 ( 29 ( 29 ( 29 J. 1608 m 6895 . N/m 5695 m/s 508 . 1 kg 225 2 2 1 2 2 1 g el = + + = + + = U U K E Let this be point 1. Let point 2 be where the beam has moved upward a distance d and where = v . ( 29 2 1 2 2 1 2 . m 6985 . E E mgd d k E = +-= gives m 7275 . = d . At this end point of motion the spring is compressed 0.7275 m – 0.6895 m =0.0380 m. At the new equilibrium position the spring is stretched 0.3879 m, so the new amplitude is 0.3789 m + 0.0380 m = 0.426 m. Energy calculations show that v is also zero when the beam is 0.426 m below the equilibrium position....

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