help me answer these 3 questions

Question

1. A 0.356 kg object connected to a light spring with a spring constant of 30.0N/m oscillates on a frictionless horizontal surface.

(a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of motion is 4cm. 

(b) What is the velocity of the object when the displacement is 2.50cm.  

C) compute the kinetic and potential energy of the system when the displacement is 1.75cm.

Answer 1

a) The total energy of a simple harmonic oscillator at any time is the sum of its potential and kinetic energies at a given time and is a constant 1/2 mv² + 1/2kx² = a constant. The total energy (the constant) is given by the potential energy at maximum displacement

1/2* 30.0 N/m *0.04² m² = 0.024 joules.

b) From the first equation given, the potential energy when the displacement is 2.5 cm is

1/2 * 30.0 N/m * 0.025² m² = 0.009375 j. Therefore the velocity at that time is obtained by solving for v in the first equation 1/2 *m*v² + 0.009375 = 0.024 = 1/2 * 0.356 kg * v² + 0.009375 = 0.024. After some algebraic manipulation, v = 0.287 m/s.

c) Again use the first equation to find the potential energy at 1.75 cm displacement

=1/2 *30*.0175² = 0.00459 j. To find the KE, substract the PE (0.00459) from the total energy

0.024 j to get KE = 0.0194 j