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About LinkBacksA particle in SHM about O has period T. It passes through P with V moving away from O. Prove that the time taken before it returns to P is given by T/pi * arctan[vT/2pix] where x is the distance OP.
In the problem, the required time is double the time taken by the particle to move from the extreme position to the point P.Originally Posted by Sunyata
The displacement x is given as
x = A*cos(ωt)
The velocity v is given by
$\displaystyle V = \omega\sqrt{A^2 - x^2}$
$\displaystyle V = \omega\sqrt{A^2 - {Acos{\{omegat}^2}$
V = ωΑ*sin(ωt)
V/x = ω*tan(ωt)
ωt = arctan(V/ωx)
t = 1/ω*arctan(V/ωx)
T' = 2t = 2/ω*arctan(V/ωx)
Put ω = 2π/T and find T'
In the problem, the required time is double the time taken by the particle to move from the extreme position to the point P.
The displacement x is given as
x = A*cos(ωt)
The velocity v is given by
V = ωsqrt{A^2 - x^2}
V = ωsqrt{A^2 - [Acos(ωt)]^2}
V = ωΑ*sin(ωt)
V/x = ω*tan(ωt)
ωt = arctan(V/ωx)
t = 1/ω*arctan(V/ωx)
T' = 2t = 2/ω*arctan(V/ωx)
Put ω = 2π/T and find T'
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