一彈簧吊在天花板下端掛一物其質量m

2013-01-27 12:10 am

一彈簧吊在天花板下端掛一物其質量m 平衡時彈簧伸長至A點且伸長量x0 若將彈簧再拉長x0/2至B點然後釋手令其上下震動(g為重力場強度) 則

1.彈力係數為何

2.彈簧的週期為何

3.振動期間物體最高點的位置在原長下方何處

4.物體所受彈力最大為多少mg

5.振動時加速度最大為何

更新1:

1. kx0=mg k=mh/x0 2. T=2拍*根號x0/g 4. k*3/2x=3mg/2 5. 3mg/2-mg=ma a=g/2

回答 (1)

天同

2013-01-27 1:03 am

✔ 最佳答案

1. Bu Hooke's law, k(xo) = mg where k is the spring constant
k = mg/(xo)

2. When the mass performs SHM(簡諧運動)
Period T = 2.pi.[square-root(m/k)] = 2.pi.square-root[m.(xo)/mg] = 2.pi.square-root[(xo/g)]

3. Since amplitude of oscillation = (xo/2)
distance below point of natural length = (xo - xo/2) = xo/2

4. The max spring force is experienced at point B,
Force = k.(Xo+xo/2) = 3k(xo)/2 = 3(mg/xo).(xo)/2 = 3mg/2

5. Max acceleration occurs at B.
Angular fequency w = square-root[k/m] = square-root[(mg/xo)/m] = square-root[g/xo]
Max acceleration = w^2.A where A is the amplitude of oscillation
Max acceleration = (g/xo).(xo/2) = g/2

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一彈簧吊在天花板 下端掛一物其質量m

回答 (1)

一彈簧吊在天花板下端掛一物其質量m