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Hrishant Goswami Grade: 10
        A particle is subjected to two simple harmonic motions given by 

x1 = 2.0 sin(100 π t) and x2 = 2.0 sin(120 π t + π/3),
Where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025
3 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
										Sol. x1 = 2 sin 100 πt

x2 = w sin (120πt + π/3)
So, resultant displacement is given by,
x = x1 + x2 = 2 [sin (100πt) + sin (120πt + π/3)]
a) At t = 0.0125s,
x = 2 [sin (100π × 0.0125) + sin (120π ×0.0125 + π/3)]
= 2 [sin 5π/4 + sin (3π/2 + π/3)]
= 2 [(–0.707) + (–0.5)] = – 2.41cm.
b) At t = 0.025s.
x = 2 [sin (100π × 0.025) + sin (120π ×0.025 + π/3)]
= 2 [sin 5π/2 + sin (3π + π/3)]
=2[1+(–0.8666)] = 0.27 cm.
3 years ago
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