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Q.

A point particle of mass 0.1 kg is executing SHM of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 18×10-3 J. The equation of motion of this particle when the initial phase of oscillation is 45° can be given by -

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a

y=0.1cos6t+π4

b

y=0.1sin6t+π4

c

y=0.4sint+π4

d

y=0.2sinπ2+2t

answer is B.

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Detailed Solution

Given : A=0.1 m, m=0.1 kg, KEmax=18×10−3 J, ϕ=π4KEmax=12mω2A2⇒ω=2KEmaxmA2=6rad/s∴ Equation of SHM is y=0.1sin(6t+π4)

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