A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8\times10^{–3} Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45°.

Correct Answer is : y=0.1sin(±4t+π4)

y=0.1\sin(\pm4t+ \frac{\pi}{4})

y=0.2\sin(\pm4t+ \frac{\pi}{4})

y=0.1\sin(\pm2t+ \frac{\pi}{4})

y=0.2\sin(\pm2t+ \frac{\pi}{4})

BITSAT 2016 Paper

Section: Physics

Question : 36 of 150

Marks: +1, -0

A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8×10–3 Joule. Obtain the equation of motion of this particle if this initial phase of oscillation is 45°.

y=0.1‌sin(±4t+‌‌
π
4
)
y=0.2‌sin(±4t+‌‌
π
4
)
y=0.1‌sin(±2t+‌‌
π
4
)
y=0.2‌sin(±2t+‌‌
π
4
)

Validate

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