Question

Un corp cu masa m=100g este supus simultan oscilatiilor paralele y1=4sin(4πt+π/6) (cm) si y2=6sin(4π+π/4) (cm). Sa se afle:

a) amplitudinea miscarii oscilatorii rezultante

b) ecuatia miscarii rezultante

c) energia oscilatorului rezultant

Answer 1

y₁=A₁sin(ωt+φ₁)

y₁=4sin(4πt+π/6) (cm)

y₂=A₂sin(ωt+φ₂)

y₂=6sin(4π+π/4) (cm)

A₁=4 cm ; A₂=6 cm

ω=4π rad/s  ; φ₁=π/6 rad   ; φ₂=π/4 rad

a) Δφ=φ₁-φ₂=π/6-π/4=(2π-3π)/12=-π/12

A²=A₁²+A₂²+2A₁A₂cosΔφ=4²+6²+2*4*6*cos(-π/12)=16+36+48[√(2+√3)]/2=52+24√(2+√3)=>A=√[52+24√(2+√3)] cm

b) tgφ=(A₁sinφ₁+A₂sinφ₂)/(A₁cosφ₁+A₂cosφ₂)=(4*0,5+6*0,7)/(4*0,866+6*0,7)=6,2/7,664=0,8=>φ=arctg(0,8)

y=Asin(ωt+φ)=√[52+24√(2+√3)]sin(4π+arctg(0,8)) (cm)

c) m=100 g=0,1 kg

A=√[52+24√(2+√3)] cm=√[52+24√(2+√3)]*10⁻² m

k=mω²=0,1*(4π)²=0,1*16*π²=1,6*10=16 N/m

E=kA²/2=16*(52+24√(2+√3)*10⁻⁴/2=8*(52+24√(2+√3)*10⁻⁴ J