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.

Results 1 to 3 of 3

- May 8th 2010, 05:59 AM #1

- Joined
- Nov 2009
- Posts
- 54

- May 8th 2010, 07:07 PM #2

- Joined
- Jun 2009
- Posts
- 806
- Thanks
- 4

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

$\displaystyle V = \omega\sqrt{A^2 - x^2}$

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

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'

- May 8th 2010, 07:11 PM #3

- Joined
- Jun 2009
- Posts
- 806
- Thanks
- 4

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'