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 J
B)0.646 J
C)0.633 J
D)0.955 J
E)0.000 J

Question

Quizplus · Accepted Answer

The total mechanical energy in simple harmonic motion is given by $ E = \frac{1}{2} k A^2 $, where $ k $ is the spring constant and $ A $ is the amplitude. The spring constant $ k $ can be found using $ k = (2\pi f)^2 m $, where $ f $ is the frequency and $ m $ is the mass. Substituting the given values, we find $ k \approx 147.9 \, \text{N/m} $. Then, $ E = \frac{1}{2} \times 147.9 \times (0.075)^2 \approx 0.646 \, \text{J} $.

A 153-Kg Piece of Iron Is Hung by a Vertical Ideal