A particle is subjected to two simple harmonic motions given by x1-2-0 sin 100 - t and x2-2-0 sin 120 - t- - 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-

Given are the equations of motion of a particle: nbsp;x1 = 2.0sin100 #960;t x2=2.0sin120 #960;t+ #960;3 The resultant displacement x will be, x = x1 + ...

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Standard XII

Physics

Projectile Time, Height and Range

A particle is...

Question

A particle is subjected to two simple harmonic motions given by x₁ = 2.0 sin (100π 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.

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Solution

Given are the equations of motion of a particle:
x₁ = 2.0sin100 $π$ t
$x_{2} = 2.0 \sin (120 π t + \frac{π}{3})$

The resultant displacement $(x)$ will be,
x = x₁ + x₂
$= 2 [\sin (100 π t) + \sin (120 π t + \frac{π}{3})]$

(a) At t = 0.0125 s
$x = 2 [\sin (100 π \times 0.0125) + \sin (120 π \times 0.0125 + \frac{π}{3})] = 2 [\sin (\frac{5 π}{4}) + \sin (\frac{3 π}{2} + \frac{π}{3})] = 2 [(- 0.707) + (- 0.5)] = 2 \times (- 1.207) = - 2.41 cm$

(b) At t = 0.025 s
$x = 2 [\sin (100 π \times 0.025) + \sin (120 π \times 0.025 + \frac{π}{3})] = 2 [\sin (\frac{10 π}{4}) + \sin (3 π + \frac{π}{3})] = 2 [1 + (- 0.866)] = 2 \times (0.134) = 0.27 cm$

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A particle is subjected to two simple harmonic motions given by x1 = 2.0 sin (100π 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.

A particle is subjected to two simple harmonic motions given by x₁ = 2.0 sin (100π 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.