Physics
A 0.70 kg mass at the end of a spring vibrates 4.0 times per second with an amplitude of 0.15 m.
Determine the velocity when it passes the equilibrium point.
Determine the velocity when it is 0.11 from equilibrium.
Determine the total energy of the system.
Reply
Answers
The velocity when it passes the equilibrium point is 0 m/s, since velocity is defined as the rate of change of position. At the equilibrium point, the position does not change. The velocity when it is 0.11 m from equilibrium can be determined using the equation for simple harmonic motion v=Aωcos(ωt). We can substitute in the values A=0.15 m, ω= 4.0 s-1, t = 0.11 m/0.15 m = 0.733 radians. This gives v = (0.15 m)(4.0 s-1)cos(0.733) = 2.08 m/s. The total energy of the system is the sum of potential and kinetic energy. The potential energy is given by Ep = ½kx2 = (½)(4π2s-2)(0.15 m)2 = 0.94 J. The kinetic energy is given by Ek = ½mv2 = (½)(0.70 kg)(2.08 m/s)2 = 0.68 J. Therefore, the total energy is Ep + Ek = 1.62 J.
