230问答网 > 某同学在资料上发现弹簧振子的周期公式为T＝2πmk，弹簧的弹性势能公式为Ep＝12kx2（式中k为弹簧的劲度系

某同学在资料上发现弹簧振子的周期公式为T＝2πmk，弹簧的弹性势能公式为Ep＝12kx2（式中k为弹簧的劲度系

2025-03-31 18:31:49
推荐回答（1个）
回答1：

（1）滑块离开平衡位置以后，第二次到达平衡位置的时间间隔为一个周期．由图乙可知，振动周期的大小为2T0．
故该系统的振动周期的大小为2T0．
    （2）由T=2π

M+m
K

得：K＝
4π2(M+m)
T2

因为Ep＝
1
2
kx2，振幅为A，所以最大弹性势能为EP＝
1
2
KA2，将K值代入
得：EP＝
2(M+m)A2π2
T2

故系统振动过程中弹簧的最大弹性势能为
2(M+m)A2π2
T2
．

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