2d. If you evaluate the position function at t= 0 s the cosine of zero is 1 so you get the maximum amplitude of the oscillation, A. This occurs at the maximum stretch of the spring from its equilibrium position. For a mass on a spring this corresponds to the maximum spring potential energy, and because the velocity is zero at the endpoints of the oscillation, the kinetic energy at the endpoints is zero, so this value represents the total energy of the system: U(x) = ½ k x² so U(x=A) = ½ k A² If the spring constant k= 19.74 N/m, what is the total energy of this system? 2e. The maximum velocity occurs at the equilibrium position when the spring is not stretched. This position is x = 0, here the spring potential energy is zero and the kinetic energy is maximum and equal to the total energy of the system. What is the maximum speed of the mass if m = 2.00 kg? Hint: the kinetic energy is given by K=½ mv and Kmax = Umax = ½ k A²

Please ONLY answer 2d and 2e

2a. A mass moving back and forth on a spring obeys the sinusoidal equation x(t) = A cos(2πft), where A is the amplitude of the oscillation or the maximum displacement of the mass from its equilibrium position, f is the frequency of the oscillation and t is the time. If the amplitude is 10.0 cm and the frequency is f = 0.500 Hz, what is the position of the mass at t = 3.00 s? 

2b. The velocity of the mass is also a sinusoidal function with the same period and frequency: v(t) = -vmax sin(2πft). Calculate vmax and the velocity of the mass v(t) when t = 3.00 s. 

2c. What is the period T of this oscillation? 

2d. If you evaluate the position function at t = 0 s the cosine of zero is 1 so you get the maximum amplitude of the oscillation, A. This occurs at the maximum stretch of the spring from its equilibrium position. For a mass on a spring this corresponds to the maximum spring potential energy, and because the velocity is zero at the endpoints of the oscillation, the kinetic energy at the endpoints is zero, so this value represents the total energy of the system: U(x) = 1⁄2 k x2 so U(x=A) = 1⁄2 k A2 If the spring constant k = 19.74 N/m, what is the total energy of this system?

2e. The maximum velocity occurs at the equilibrium position when the spring is not stretched. This position is x = 0, here the spring potential energy is zero and the kinetic energy is maximum and equal to the total energy of the system. What is the maximum speed of the mass if m = 2.00 kg? Hint: the kinetic energy is given by K = 1⁄2 mv2 andKmax =Umax =1⁄2kA2