. Two parallel S.H.M.s represented by
x1
= 5sin (4π t + 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.
[Ans: 7.936 cm, 54° 23']
Answers
Answer:
Two simple harmonic motions are represented by y1=4sin(4πt+π/2) and y2=3cos (4πt). What is the resultant amplitude
Given : Two parallel S.H.Ms represented by x₁ = 5sin(4πt + π/3) cm and x₂ = 3sin(4πt + π/4) cm are superposed on a particle.
To find : The amplitude and epoch of the resultant SHM.
solution : using superposition principle,
amplitude of resultant is given by,
A = √{A₁² + A₂² + 2A₁A₂cosΦ}
here, A₁ = 5, A₂ = 3 and Φ = π/3 - π/4 = π/12
now A = √{5² + 3² + 2(5)(3)cos(π/12)}
= √(25 + 9 + 30 × 0.966)
= √(62.98)
= 7.936 cm
epoch of the resultant, θ = tan¯¹ [(A₁sinΦ₁ + A₂sinΦ₂)/(A₁cosΦ₁ + A₂cosΦ₂)]
= tan¯¹ [(5sin(π/3) + 3sin(π/4))/(5cos(π/3) + 3cos(π/4))]
= tan¯¹[ (5 × √3/2 + 3 × 1/√2)/(5 × 1/2 + 3 × 1/√2)]
= tan¯¹ [ (5√3 + 3√2)/(5 + 3√2)]
= 54° 23'
Therefore amplitude is 7.936 cm and epoch of the resultant is 54°23'