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
1
Expert's answer
2018-09-20T10:55:09-0400
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
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