A particle of mass 200 g executes simple harmonic motion along a straight line. If the maximum potential energy of the particle during its oscillation is 0.9 J, what is its speed when it crosses the mean position?

Step-by-step solution

Mass of the particle, m = 200 g = 0.2 kg Maximum potential energy, U_ max = 0.9 J In simple harmonic motion, the maximum kinetic energy is equal to the maximum potential energy. K_ max = U_ max = 0.9 J The kinetic energy is maximum at the mean position, where the speed is v_ max . 1 2 m v_ max ^2 = 0.9 1 2 0.2 v_ max ^2 = 0.9 0.1 v_ max ^2 = 0.9 v_ max ^2 = 9 v_ max = 3.0 m/s Answer: 3.0 m/s