Homework 11 Solution

The minimum period is therefore m b t2 k k2 1 21 1 2

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Unformatted text preview: nic (12) with periods l h = √ ≈ 0.29l (13) 23 m(k 1 k 2) (a) T2 1k 2 A quick second derivative test or ak plot of dT/dh verifies that this is indeed a minimum, not a √ √ maximum. The minimum period is therefore m (b) T2 k k2 1 21 1 2 12 l + 12 l = 2π = 2π Tmin = T l l g 2√3 h= √ 23 k1 l √ ≈ 2.26 s 3g k2 4. A block of mass m is connected to two springs of force constants k1 and k2 as shown below. The m block moves on a frictionless table after it is displaced from equilibrium and released. Determine the period of simple harmonic motion. (Hint: what is the total force on the block if it is displaced by an amount x? (a) ly to M as isk is sk is bly is ition f the s k1 k2 m (b) Solution: Say we displace the block to the right by an amount x. Both springs will try to bring Figure P15.71 the block back toward equilibrium - one will pull, one will push, but both will act in the same direction. That means the net force is 72. A lobsterman’s buoy is a solid wooden cylinder of radius r Fnet M It s w k2 h = − (k1 + 2 ) d so th and mass = .−k1ix −eigx ted at onekenx = maat it flo...
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This note was uploaded on 01/26/2014 for the course PH 105 taught by Professor Stanjones during the Fall '08 term at Alabama.

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