Se da:
m=2kg
y=4*[sin(20t)+√3*cos(20t)]
A=?
φ0=?rad
Rezolvare:
Aducem ecuatia data, intr-o forma obisnuita[y=A*sin(wt+φ0)], stiind ca
sin(a+b)=sina*cosb+sinb*cosa
4*[sin(20t)+√3*cos(20t)]=8*[1/2*sin(20t)+√3/2*cos(20t)]
1/2=cos(60)=cos(π/3)
√3/2=sin(60)=sin(π/3)
8*[1/2*sin(20t)+√3/2*cos(20t)]=8*[cos(π/3)*sin(20t)+sin(π/3)*cos(20t)]=
=8*sin(20t+π/3)
y=8*sin(20t+π/3), de unde:
A=8, φ0=π/3rad.
