Question

Two parallel S.H.M.s represented by x₁ = 5sin (4πt + π/3) cm and x₂ = 3sin (4πt + π/4) cm are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M.

Answer 1

Given : x₁ = 5sin (4πt + π/3) = A₁ sin(ωt + α),

x₂ = 3sin (4πt + π/4) = A₂ sin(ωt + β),

∴ A₁ =5 cm, A₂ =3 cm, α = π/3 rad, β = π/4 rad

(i) Resultant amplitude,

R = \(\sqrt{A_1^2+A_2^2+2A_1A_2cos(α-β)}\)

∴ R = \(\sqrt{5^2+3^2+2(5)(3)cos(π/3-π/4)}\)

= \(\sqrt{25+9+30cos(π/3-π/4)} = \sqrt{62.98}\)

= 7.936 cm

(ii) Epoch of the resultant SHM,

δ = \(tan^{-1}(\frac{A_1sinα+A_2sinβ}{A_1cosα+A_2cosβ})\)

= \(tan^{-1}(\frac{5sin(π/3)+3sin(π/4)}{5cos(π/3)+3cos(π/4)})\)

= \(tan^{-1}(\frac{5(0.866)+3(0.7071)}{5(0.5)+3(0.7071)})\)

= \(tan^{-1}(1.396)\) = 54⁰23’

N-m