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

Answer 1

E = 0.645J

Step-by-step explanation:

In order to calculate the total mechanical energy of the system, you take into account that if the zero of energy is at the equilibrium position, then the total mechanical energy is only the elastic potential energy of the spring.

You use the following formula:

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

k: spring constant = ?

A: amplitude of the oscillation = 7.50cm = 0.075m

The spring constant is given by:

`f=(1)/(2\pi)\sqrt{(k)/(m)}`

`k=4\pi^2f^2m` (2)

f: frequency of the oscillation = 1.95Hz

m: mass of the piece of iron = 1.53kg

You replace the expression (1) into the equation (2) and replace the values of all parameters:

`E=(1)/(2)(4\pi^2f^2m)A^2=2\pi^2f^2mA^2\\\\E=2\pi^2(1.95Hz)^2(1.53kg)(0.075m)^2=0.645J`

The totoal mechanical energy of the system is 0.645J