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calgal260
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Homework Statement
A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth in simple harmonic motion on a frictionless horizontal surface. At one extreme end of the oscillation cycle, where the blocks come to a momentary halt before reversing the direction on their motion, the top block is suddenly lifted vertically upward, without changing the zero velocity of the bottom block. The simple harmonic motion then continues. What happens to the amplitude and the angular frequency of the ensuing motion?
Homework Equations
PE(elastic)= 0.5kA^2
The Attempt at a Solution
The answer that they give us is amplitude remains the same, while angular frequency increases.
I understand the angular frequency part but not the amplitude part. When the spring sets in motion and you look at the total energy equation, shouldn't the elastic potential energy be less now, now that angular frequency is no longer zero (since it's not halted) and you have rotational and translational kinetic energy? And they tell you that there is no friction, which means energy is conserved. How can energy be conserved when the potential elastic energy (from which the amplitude is derived) remains the same, and thus amplitude as well?
Here is the book explanation. It still confuses me:
At the instant the top block is removed, the total mechanical energy of the remaining system is all elastic potential energy and is 12 kA2 (see Equation 10.13), where A is the amplitude of the previous simple harmonic motion. This total mechanical energy is conserved, because friction is absent. Therefore, the total mechanical energy of the ensuing simple harmonic motion is also 12 kA2 , and the amplitude remains the same as it was previously. The angular frequency ω is given by Equation 10.11 as ω = mk . Thus, when the mass m attached to the spring decreases, the angular frequency increases.