A particle executes shm with an amplitude of 10 cm and a period of 5 s. Find the velocity and acceleration of the particle of a distance 5 cm from the equilibrium position.

Explanation

`=v=omegasqrt(a^2-y^2)=(2pi)/Tsqrt(a^2-y^2)`
`= (2xx3.14)/5 sqrt(10^2-5^2)=10.88 cm//s`
`= 0.108m//s`
Acceleration ` = a =- omega^2y=(4pi^2)/(T^2)y`
`= - 4 xx3.14^2xx5`
`= 197.2 cm//s^2 = 1.97 m//s^2`