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 90^ .

Step-by-step solution

The resultant amplitude A of two simple harmonic motions of amplitudes A₁ and A₂ with a phase difference is given by: A = A₁^2 + A₂^2 + 2A₁A₂ Given A₁ = 3.0 cm , A₂ = 4.0 cm , and = 90^ . Substituting the values: A = (3.0)^2 + (4.0)^2 + 2(3.0)(4.0) 90^ Since 90^ = 0 , we get: A = 9.0 + 16.0 A = 25.0 = 5.0 cm