Answer to Question #81076 in Mechanics | Relativity for Eric Graham

Question

A 280g object attached to a horizontal spring moves in simple harmonic motion with a period of 0.120s. The total mechanic energy of the spring-object system is 2.00J
What is the spring constant

Answer 1

Mass:
m=0.28kg
Period:
T=0.12s
Total mechanic energy:
E=2J
Solution:

The oscillation period of a spring pendulum can be calculated by the following formula:
T=2π√(m/k)
where k is spring stiffness (spring constant):
k=4π^2 m/T
k=4*〖3.14〗^2*0.28/〖0.12〗^2 =767.6 N/m
Thus, data on the total mechanical energy of the spring-object system - are redundant, so they are not involved in the calculation