The displacement of a particle executing SHM is given by x=0.10sin ((2pit)/T+phi) where all the quantities are in SI units. The time period of vibration is 3s. At t=0, the particle is displaced by 0.05 m. Find (i) the initial phase (ii) the phase angle corresponding to 0.0866 m and (ii) the phase difference between any two positons of the particle 2s apart.

The displacement of a particle executing SHM is given by x=0.10sin ((2

1.	The displacement of a particle executing SHM is given by x=0.10sin ((2pit)/T+phi) where all the quantities are in SI units. The time period of vibration is 3s. At t=0, the particle is displaced by 0.05 m. Find (i) the initial phase (ii) the phase angle corresponding to 0.0866 m and (ii) the phase difference between any two positons of the particle 2s apart.
Answer» Solution :For the particle in SHM `x=0.10 sin ((2pit)/T+PHI)` `=0.10sin((2pi)/3t+phi)( :. T=3s)` (i) At `t=0, x=0.05m` `0.05=0.10sin ((2pi)/3xx0+phi)=0.10sin phi` `sin phi =1/2impliesphi=(pi)/6`radian or `30^(@)` (ii) Substituting `x=0.0866`m in `x=0.10sin ((2pi)/3t + phi)` `0.0866=0.10sin((2pi)/3t+phi)` `sin((2pi)/3t+phi)=0.0866/0.10=0.866` `:.` Phase ANGLE `((2pi)/3t+phi)=(pi)/3` radian or `60^(@)`. (iii) Phase angle at `t=t_(1)` is `phi_(1)=((2pi)/3xxt_(1)+phi)` and at `t=t_(2)` is `phi_(2)=((2pi)/3xxt_(2)+phi)` Phase difference `=phi_(2)-phi_(1)` `=((2pi)/3t_(2)+phi)-((2pi)/3t_(1)+phi)=(2pi)/3(t_(2)-t_(1))` `=(2pi)/3x2-(4pi)/3` radian `( :. t_(2)-t_(1)=2)`