A particle executes shm with a period 8 s. Find the time in which half the total energy is potential.

A particle executes shm with a period 8 s. Find the time in which half

1.	A particle executes shm with a period 8 s. Find the time in which half the total energy is potential.
Answer» Solution :T = 8s Total energy `= 1/2 momega^2a^2` POTENTIAL energy`=1/2 m OMEGA^2y^2` where y is the DISPLACEMENT of the PARTICLE from the equilibrium position Given that `1/2 XX`Total energy = Potential energy `1/2[1/2momega^2a^2]=1/2momega^2y^2` `1/2a^2=y^2` `y=a/sqrt2` `omega=(2pi)/T=(2pi)/8=pi/4` `y= a sin omegat= a sin""(pi)/4t` `a/sqrt2 = a sin""pi/4t` `sin""pi/4 = sin""pi/4t` `pi/4 = pi/4t` t = 1 s