A 0.187 kg mass is attached to a spring and executes simple harmonic motion with a period of 0.27 s. the total energy of the system is 2.5 J. Find the force constant of the spring.
2007-11-20 15:20:23 · 2 answers · asked by Yami 1 in Science & Mathematics ➔ Physics
The period of a mass on a spring T = 2 pi sqrt ( m / k ).
Plug in T = .27 s and m = .187 kg and solve for k.
2007-11-20 15:44:48 · answer #1 · answered by jgoulden 7 · 1⤊ 1⤋
If you use energy = mv^2/2 answer D is correct IFF the doubling is done while velocity v is maximum. This doubles the kinetic energy, which at that time is equal to the total energy. If you use energy = kx^2/2 answer A is correct IFF the doubling is done while displacement x is maximum. This doubles the potential energy, which at that time is equal to the total energy. Doubling done at other than the respective maximum still doubles the respective type of energy, but that energy is less than the total energy so total energy doesn't double.
2016-04-05 01:12:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋