Question

Un oscilator efectueaa 180 oscilatii intr-un minut si are A=2cm , cu omega = 2*3,14/3 si frecventa= 3 Hz.Sa se afle

a) ecuatia x(t) daca f0= 3.14(PI)/12
b) Vm si Am.

Answer 1

x(t)=2sin(2*3,14/3*t + 3.14/12) (cm)

Vmax=ω*A=(2*3,14/3)*0.02=0.04*3.14/3

amax=ω^2*A=0.08*3.14^2/3