Work, Power and EnergyHard
Question
A point particle of mass 0.1 kg is executing S.H.M. of amplitude of 0.1 m. When theparticle passes through the mean position, its kinetic energy is × 10-3 Joule. Obtain the equation of motion of this particle if this intial phase of oscillation is 45o.
Options
A.y = 0.1 sin 
B.y = 0.2 sin
C.y = 0.1sin 
D.y = 0.2 sin
Solution
The displacement of a partical in S.H.M. is given by
y = a sin(ωt + φ)
velocity =
= ω a cos(ωt + φ)
The velocity is maximum when the particle passes through the mean position i.e.,
=
= ωa he kinetic energy at this instant is given by
= ω2a2 = 8 × 10-3 joule
or
× (0.1) ω2 × (0.1)2 = 8 × 10-3
Solving we get ω = +4
Substituting the values of a ω and φ in the equation of S.H.M we get
y = 0.1 sin (+4 + π/4)
y = a sin(ωt + φ)
velocity =
= ω a cos(ωt + φ)The velocity is maximum when the particle passes through the mean position i.e.,
=
= ωa he kinetic energy at this instant is given by = ω2a2 = 8 × 10-3 jouleor
× (0.1) ω2 × (0.1)2 = 8 × 10-3Solving we get ω = +4
Substituting the values of a ω and φ in the equation of S.H.M we get
y = 0.1 sin (+4 + π/4)
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