解：A=0.24mw= 2元 2元 兀 T 4 2 =1.57s1 xo=A=0.24m t=0 振动方程为：x=0.24cos2 t=0.5s x=0.24c0s2×0.5) =0.24c0S0.25π 2 =0.24× 2 =0.17m 结束返回 t =0 ω 2π T = 4 =1.57s-1 = 2 = 2π π v 0 = 0 x 0 = A =0.24m x =0.24 cos t 2 π t =0.5s x =0.24 cos( × ) 2 π 0.5 =0.24 cos0.25π × =0.17m 2 2 =0.24 振动方程为： 解：A=0.24m φ = 0 结束 返回