Question

In a horizontal spring-mass configuration, the position of the 0.44 kg mass is described by [x = (0.49 m) sin(2.39t)]

a) What is in meters the amplitude of these oscillations?

b) What is in m/s the maximum speed of the mass under these conditions?

c) What is in Joules the spring potential energy stored in this system when the spring is stretched at a maximum?

d) What is in Joules the kinetic energy of the mass when the spring is not stretched?

e) What is in Joules the kinetic energy of the mass when the spring is stretched by 0.16 m?

Answer 1

Mass m = 0.44 kg

Position x = 0.49 sin 2.39 t

Compare this with x = A sin wt   you get

A = 0.49 m

w = 2.39 rad / s

a) The amplitude of these oscillations A = 0.49 m

b) The maximum speed of the mass under these conditions V = Aw

                          V = 1.1711 m / s

c) The spring potential energy stored in this system when the spring is stretched at a maximum

               U = ( 1/2) K A ²

Where K = Spring constant = mw ²

             = 2.513 N / m

So, U = 0.3017 J

d) The kinetic energy of the mass when the spring is not stretched K.E = U

Since from law of conservation of energy .

So, K.E = 0.3017 J

e) The kinetic energy of the mass when the spring is stretched by 0.16 m is KE ' = KE - U '

Where U ' = Potential enery of the spring when it streached by 0.16 m

                = ( 1/ 2) Kx ²     where x = 0.16 m

               = 0.03216 J

So, KE ' = 0.3017 J - 0.03216 J

            = 0.2695 J