Question

一簡(jiǎn)諧振子沿x軸振動(dòng),平衡位置在坐標(biāo)原點(diǎn).t=0時(shí)刻振子的位移x=-0.1 m;t= s時(shí)刻x=0. 1 m;t=4 s時(shí)刻x=0.1 m.該振子的振幅和周期可能為(    )

A.0.1 m, s          B.0.1 m, 8 s

C.0.2 m, s          D.0.2 m, 8 s

 

Answer 1

【答案】

ACD

【解析】本題考查簡(jiǎn)諧運(yùn)動(dòng)的周期性和對(duì)稱性.設(shè)振子在AOA′之間運(yùn)動(dòng),B和B′關(guān)于O對(duì)稱,且距平衡位置的距離為0.1 m,如圖所示.由題意,t=0時(shí)在B點(diǎn),當(dāng)振子由B向右運(yùn)動(dòng),t_BB′= s,t_B′A′B′=(4--nT) s,則=t_OB′+t_B′A′= (t_BB′+t_B′A′B′)=2 s,所以n=0時(shí),T=8 s; n=1時(shí),T= s,振幅A>0.1 m,則CD正確;當(dāng)振子由B向左運(yùn)動(dòng),t_B′A′B′= s,而t_BABB′= s,B在AO之間,根據(jù)運(yùn)動(dòng)的對(duì)稱性,應(yīng)該大于,此情景不符;當(dāng)A和B重合,當(dāng)振子由B開(kāi)始運(yùn)動(dòng),則T_BO+T_OB′= s,從B′出發(fā)再回到B′,恰好一個(gè)周期,則+T=4 s,所以T= s,恰好符合情景,則A正確.