👤
Mycuta
a fost răspuns

sa se determine faza initiala de oscilatie a unui oscilator arminic liniar, daca la momentul t=T/12, oscilatorul se afla la elgatia y=√3*A/2
VA ROG!
​

Răspuns :

Baiatul122001

y(t)=Asin(ωt+φ₀) (m)

ω=2π/T=>T=2π/ω

t=T/12=2π/ω/12=π/6ω

y=A√3/2

y(t)=Asin(ωt+φ₀)<=>A√3/2=Asin(π/6+φ₀)<=>sin(π/6+φ₀)=√3/2=>π/6+φ₀=arcsin√3/2=>π/6+φ₀=π/3=>φ₀=π/3-π/6=(2π-π)/6=π/6=>φ₀=π/6 radiani