A particle of mass 0.2kg is executing SHM along x-axis with a frequency of `25/piHz`. If its kinetic energy is 0.5 J at x = 0.04 m, then find its amplitude of vibration?

Explanation

Here, `n=25/piHz, "so",omega=2pi*25/pi=50"rad"*s^(-1)`
We know, `K=1/2momega^(2)(A^(2)-x^(2))`
or, `(2K)/(momega^(2))=A^(2)-x^(2)`
or, `A^(2)=(2K)/(momega^(2))+x^(2)=(2xx0.5)/(0.2xx(50)^(2))+(0.04)^(2)=0.0036`
or, A = 0.06m.