The displacement of a particle executing SHM is given by x = 0.10 sin ((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 0.05m. Findthe phase angle corresponding to 0.0866 m

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

1.	The displacement of a particle executing SHM is given by x = 0.10 sin ((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 0.05m. Findthe phase angle corresponding to 0.0866 m
Answer» Solution :For the particle in SHM, the displacement , `x= 0.10sin((2pit)/(T)+phi)` `= 0.10sin((2pi)/(3)t+phi)( :.T=3s)` substituting x= 0.0866m in `x= 0.10sin((2pi)/(3)t+phi)` `0.0866= 0.10 sin ((2pi)/(3)t+phi)` `sin((2pi)/(3)t+phi)=(0.0866)/(0.10)= 0.866` `:.`PHASE angle `((2pi)/(3)t+phi)= (PI)/(3)` radian or `60^(@)`