n 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 metersSHM :  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