Question

一道物理题！！！！

一竖直悬挂的弹簧振子，将振子下拉使弹簧伸长3mg/k,然后有精止释放，要使振子动能达到(mg)^2/k,至少需要经历的时间是多少？？谢谢了！！！

Answer 1

将振子下拉使弹簧伸长3mg/k,然后有静止释放，说明振动的总能量是
E = 1/2 k (3mg/k)^2 = 9/2 (mg)^2/k
使振子动能Ek达到(mg)^2/k时，势能为
Ep = E - Ek = 9/2 (mg)^2/k - (mg)^2/k = 7/9 [9/2 (mg)^2/k] = 7/9 E
所以
7/9 1/2 k A^2 = 1/2 k A^2 cos^2ωt
得 cosωt = √7 / 3
另外，ω = √(k/m)
所以, 至少需要经历的时间是
t = √(m/k) * arccos(√7 / 3)