A particle is subjected to two simple harmonic motions given by X_{1} = 2.0 sin(100 πt) and x_{2} = 2.0 sin(120 πt + π/3), where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0025.

(a) -2.41 cm

(b) 0.27 cm

Given equations are

=2.0sin 100ϖt , =2.0sin (120ϖt +ϖ/3)

Resultant wave will be x= = 2[sin 100ϖt + sin (120ϖt +ϖ/3)]

a) At t= 0.0125 s

X= 2[sin 100ϖ 0.0125 + sin (120ϖ × 0.0125 +ϖ/3)]

= 2[sin(5 π/4) + sin ( 3π /2 + π /3)

=2[-0.707-0.5]

= -2.41cm

b) At t= 0.025 s

X = 2[sin 100ϖ 0.025 + sin (120ϖ × 0.025 +ϖ/3)]

= 2[sin(10 π/4) + sin ( 3π + π /3)

= 2[1-0.866] =0.27cm

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