A particle is subjected to two simple harmonic motions given by X1 = 2.0 sin(100 πt) and x2 = 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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