A particle is subjected to two simple harmonic motions of the same time period in the same direction. The amplitude of the first motion is 3.0 cm while that of the second is 4.0 cm . Determine the resultant amplitude if the phase difference between the motions is 60^ .

Step-by-step solution

The resultant amplitude A of two simple harmonic motions in the same direction is given by the formula: A = A₁^2 + A₂^2 + 2A₁A₂ Substituting the given values A₁ = 3.0 cm , A₂ = 4.0 cm , and = 60^ : A = 3.0^2 + 4.0^2 + 2(3.0)(4.0) 60^ A = 9 + 16 + 24 1 2 A = 25 + 12 A = 37 6.08 cm Rounding to one decimal place, we get 6.1 cm .