Question

A 1.53-kg piece of iron is hung by a vertical ideal spring. When perturbed slightly, the system is moves up and down in simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. If we choose the total potential energy (elastic and gravitational) to be zero at the equilibrium position of the hanging iron, what is the total mechanical energy of the system?

A. 0.844 JB. 0.646 JC. 0.000 JD. 0.955 JE. 0.633 J

Answer 1

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

`E = (1)/(2) kA^2`

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:

`\omega^2 = (k)/(m)`

`k = m\omega^2\rightarrow \omega = 2\pi f` for f equal to the frequency.

`k = 1.53(2\pi 1.95)^2`

`k = 229.44`

Finally the energy released would be

`E = (1)/(2) (229.44)(0.075)^2`

`E = 0.6453J \approx 0.646J`

Therefore the correct answer is B.