Răspuns:
a)A=10*10⁻²=10⁻1m=1/10m=0,1m
pulsatia ω=π
perioada T=2π/ω=2π/π=2 s
f=1/T=1/2=,0,5hz
faza initiala=π/2
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T/3=2/3s
y(2/3)=0,1(2/3π+π/2)=0,1*sin 7π/6=0,1*(-1/2)=0,05m
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c)Ec=Ep
mω²A²/2*cos²(πt+π/2)=mω²A²/2sin²(πt+π/2)║:mω²A/2
cos²(πt+π/2)=sin²(πt+π/2) Imparti egalitatea prin cos²(πt+π/2)=>
1=tg²(πt+π/2)=.>
tg(πt+π/2)=±1
caz 1
tg(πt+π/2)=1=>
πt+π/2=π/4
πt=π/4-π/2= -π/4=>
t1= -4 nu convine problemei
caz 2
tg(πt+π/4)= -1=>
πt+π/4= -π/4
πt=-π/2
t2= -1/2
Explicație:
