equation of motion and force:
F = m a = m d²x/dt² = - k x
d²x/dt² = - k/m x
let x = A Sin (ωt + Ф)
then d²x/dt² = - A ω² Sin(ωt+Ф) = - ω² x
k = 116 N/m or 16 N/m ??? which one ?
m = 0.1 kg
A = 0.20 meters
SHM : angular frequency = ω = √(k/m) = √(116/0.1) = 34 rad/sec
if k = 16 N/m, ω = √160 = 4√10 rad/sec
x = A Sin (ωt + Ф) = displacement from the mean position
v = velocity of the particle executing the SHM
v = dx/dt = A ω Cos (ω t + Ф)
v = ω √A² - x² ) = ω A √[1 - x²/A² ]
x = 1.0 m this value is not possible, as amplitude is 0.20 m. SO x has to be less than 0.2 m. Is it 0.1 meters ? x has to be less than or equal to amplitude.
v = 4√10 rad/sec * √[0.2² - 0.1²] meters = 2.19 m/sec
Energy of the spring mass system :
= 1/2 m v² at the mean equilibrium position , as x = 0 and PE = 0
= 1/2 k A² at the extreme position when x = A, as v = 0 and KE = 0.
= 1/2 * 0.1 kg * 0.20² = 0.002 Joules