Question

sa se calculeze pentru oscilatia armonica descrisa in ecuatia y=50 sin(628 t +Pi/3)cm perioada ,amplitudinea, frecventa, faza initiala, elongatia la momentul initial si la momentele t1=1/3400 si t2=1/300

Answer 1

y=50sin(628t+π/3)  
ω=628 rad/s  
fi indice 0=π/3
 A=50cm=0,5m 

Perioada T=2π/ω  =>  T=2π/628=6,28/628=0,01 s  
T=0,01s 

Frecventa (miu) ∨=1/T  => ∨=1/0,01=100
 ∨=100Hz

Elongatia la momentul initial adica la fi indice 0=π/3
 y(t0)=Asin(ωt+fi0)
de unde
y(t0)=Asin fi indice 0)  y(t0)=0,5*sin π/3=
=0,5*√3/2=50*10 la minus 2 *√3/2=0,25*√3=0,425m
 y(t0)=0,425m

 y(t1)=Asin(ωt1+fi0) =0.5sin1,22=0,5*0.021=0,0105m 
y(t1)=0,0105m

y(t2)=0,5sin3,13=0,5*0,054=0,027m
y(t2)=0,027m

Sper sa intelegi ce am facut! :D