TS EAMCET 03 May 2019 Shift 2 Solved Paper

Section: Physics

Question : 92 of 160

Marks: +1, -0

A particle of mass 0.1 kg is executing simple harmonic motion of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10−3 J. If the initial phase is 45°, the equation of its motion is (Assume, x(t) as the position of the particle at time t)

x (t) = 0.1 sin (4t+
π
4
)
x (t) = 0.1 sin (16t+
π
4
)
x (t) = 0.1 sin (2(t+
π
4
))
x (t) = 0.1 sin (2t+
π
4
)

Validate

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