Question

A mass of 4.00kg on a spring is pulled a distance of 0.250 m from equilibrium and then released from rest. It takes 0.125s to make a first pass through the equilibrium position. a. what is the period and spring constant for this system? b. what is the total energy of the system? c. what is the kinetic energy of the system as it passes through the equalibrium position? d. where is the mass 1.5s after it is released?

Answer 1

here,

mass , m = 4 kg

x = 0.25 m

equating the forces vertically

m * g = K * x

4 * 9.81 = K * 0.25

K = 156.96 N/m

t = 0.125 s

a)

the period , T = 2 * t

T = 0.5 s

b)

the total energy of the system , TE = 0.5 * K * x^2

TE = 0.5 * K * x^2

TE = 0.5 * 156.96 * 0.25^2 = 4.91 J

c)

the kinetic energy of the system as it passes through the equalibrium position, KE = TE = 4.91 J

d)

after 1.5 s

that means after 3 time period

the position , x' = 0.25 m below the equilibrium position