To solve this problem it is necessary to apply the concepts related to the elastic potential energy from the simple harmonic movement.
Said mechanical energy can be expressed as
[tex]E = \frac{1}{2} kA^2[/tex]
Where,
k = Spring Constant
A = Cross-sectional Area
From the angular movement we can relate the angular velocity as a function of the spring constant and the mass in order to find this variable:
[tex]\omega^2 = \frac{k}{m}[/tex]
[tex]k = m\omega^2\rightarrow \omega = 2\pi f [/tex] for f equal to the frequency.
[tex]k = 1.53(2\pi 1.95)^2[/tex]
[tex]k = 229.44[/tex]
Finally the energy released would be
[tex]E = \frac{1}{2} (229.44)(0.075)^2[/tex]
[tex]E = 0.6453J \approx 0.646J[/tex]
Therefore the correct answer is B.
