Sebuah benda bermassa m=0,25 kg melakukan osilasi dengan periode 0,2 sekon dan amplitudo A= 5times 10^-2m . Pada saat simpangannya y=2times 10^-2 m, hitunglah (a)percepatan benda, (b)gaya pemulih, (c) energi potensial, dan (d) energi kinetik benda!

With that,(a) The frequency f = 1/T = 1/0.2s = 5 Hz. Acceleration a = -(2πf)²*n = −(2π5Hz)²*0.02m = -157 m/s²(b) The restoring force, F = m*a = 0.25kg*-157 m/s² = -39.25N.(c) Now, for PE, we need spring constant. k = m*(2πf)² = 0.25kg*(2π5Hz)² = 247 kg/s². So, PE = 0.5*k*n² = 0.5*247 kg/s²*0.02m² = 0.049 joules.(d) The total energy TE= 0.5m*(A^2)*(2πf)² = 0.5*0.25kg*(20 m)²*(2π5Hz)² = 3.138 joules. The kinetic energy, KE = TE - PE = 3.138 J - 0.049 J = 3.089 joules.

The given oscillation is simple harmonic.(a) The acceleration in simple harmonic motion (SHM) is given by a = −(2πf)²*n , where f is the frequency, n is the displacement and the minus shows it towards opposite of the displacement direction (pointing towards the natural equilibrium position). The frequency in "seconds" (s) is the reciprocal of the given period T.(b) The restore force F for SHM is F = ma. (c) The potential energy (PE) in SHM is PE = 0.5*k*n², where k is the spring constant. Also, k = m*(2πf)² for SHM.(d) The total energy (TE) in SHM is conserved. As, TE = sum of kinetic and potential energy.That's why PE + KE (Kinetic Energy) = 0.5m (A^2)(2πf)² (approximately), where A = amplitude. So, KE = TE − PE.