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Q.A particle executes simple harmonic motion according to the relation: y = 10 Sin(2πt + π/6) metre. Find the displacement, velocity and acceleration of the particle at t = 2 second.

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2017Subjective· 3mImportance★★★★★
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Substituting t = 2 s (and using periodicity of sine) into y, v = dy/dt and a = -ω^2 y gives y = 5 m, v ≈ 54.4 m/s and a ≈ -197.4 m/s^2.

Given: y = 10 sin(2πt + π/6) metre, so amplitude A = 10 m and angular frequency ω = 2π rad/s.

Displacement at t = 2 s:

Argument = 2π(2) + π/6 = 4π + π/6

Since sine has period 2π, sin(4π + π/6) = sin(π/6) = 1/2

y = 10 × (1/2) = 5 m

Velocity, v = dy/dt = 10 × 2π cos(2πt + π/6) = 20π cos(2πt + π/6)

At t = 2 s: cos(4π + π/6) = cos(π/6) = √3/2 ≈ 0.866

v = 20π × 0.866 ≈ 62.83 × 0.866 ≈ 54.4 m/s

Acceleration, a = dv/dt = -ω^2 y = -(2π)^2 × y = -4π^2 × y …

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