1. ## Shm again!

A 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.

2. Originally Posted by Sunyata
A 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.
The displacement x is given as
x = A*cos(ωt)
The velocity v is given by
$V = \omega\sqrt{A^2 - x^2}$

$V = \omega\sqrt{A^2 - {Acos{\{omegat}^2}$
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'

3. Originally Posted by Sunyata
A 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.
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'