0.136 s is the earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy.
A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force F proportional to the displacement x, where k is a positive constant.
frequency of system = f
time = t
angular frequency = ω
amplitude = A
spring constant = k
mass = m
The angular frequency = 2[tex]\pi[/tex]. f
= 2[tex]\pi[/tex]f
= [tex]2\pi (0.72s^{-1} )[/tex]
= 4.5216[tex]s^{-1}[/tex]
The phase constant φ is zero since the system is no longer at rest.
The elastic potential energy U of the spring must be divided by the kinetic energy K of the block.
[tex]K = \frac{1}{2} U[/tex]
[tex]tan^{2}[/tex](ωt) = 0.5
ωt = [tex]tan^{-1} (\sqrt{0.5} )[/tex]
t = 0.6155/ω
= 0.6155/4.5216 rad/s
= 0.136 s
Hence, 0.136 s is the earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy.
Learn more about kinetic energy here:
https://brainly.com/question/15764612
#SPJ4