Machanics Homework 8 - Solution

Machanics Homework 8 - Solution - Homework 8(6/2 11.1 x0 =0...

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Homework 8 (6/2) 11.1 平衡位置 x 0 =0 >0 為穩定平衡、 <0 為不穩定平衡。 (a) V ( x )= k 2 x 2 + k 2 x V ' ( x )= k x k 2 x 2 V '' ( x )= k + 2 k 2 x 3 V ' ( x 0 )= k x 0 k 2 x 0 2 = 0 x 0 = 3 k 得平衡點。 V '' ( 3 k )= k + 2 k 2 k = 3 k > 0 ---穩定平衡。 (b) V ( x )= kx e bx V ' ( x )= k e bx ( 1 b x ) V '' ( x )= k b e bx ( b x 2 ) V ' ( x 0 )= k e bx 0 ( 1 b x 0 ) = 0 x 0 = 1 b 得平衡點。 V '' ( 1 b )= k b e b 1 b ( b 1 b 2 ) =− k b e 1 < 0 ---不穩定平衡。 (c) V ( x )= k ( x 4 b 2 x 2 ) V ' ( x )= k ( 4 x 3 2 b 2 x ) V '' ( x )= k ( 12 x 2 2 b 2 ) V ' ( x 0 )= k ( 4 x 0 3 2 b 2 x 0 )= 0 x 0 = 0 or x 0 b 2 得平衡點。 V '' ( 0 )=− 2 k b 2 < 0 ---不穩定平衡。 V '' b 2 )= 4 k b 2 > 0 ---穩定平衡。 (d) 微小震盪可視為簡諧震盪,其彈性係數 κ= d 2 V dx 2 、頻率 ω= κ m = ¨ V m 、週期 T = 2 π m ¨ V 對於 (a) ω= 3 k m T = 2 π m 3 k = 2 π 3 (in cgs units) 對於 (c) ω= 4 kb 2 m T = 2 π m 4 kb 2 (in cgs units) 11.5 位能 V = mg ( h 1 + h 2 ) 球面的高度 h 1 = b cos θ 質心到球面的高度 h 2 = d cos ϕ d 是質心到球面接觸點的距離、ψ 是方塊傾斜的角度。 sin (θ−ϕ)= b θ d cos (θ−ϕ)= a d
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