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Simple Harmonic Motion
Solution 3
Practice Problems
SHM Overview
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After the block is released from x=A, it will move to the left direction until it reaches x = -A, at which point the block will begin moving to the right until it reaches x=A again, assuming no energy has been lost.
If the period is doubled, the frequency is halved.
When the block is held at x=A, all the energy is stored as potential. When it is released, the potential energy is converted into kinetic, therefore the potential energy decreases. Because U=kx^2, a decrease in potential energy implies a decrease in extension, and we can assume that the block is moving to the left until it reaches its equilibrium point.
Energy must be conserved. The block holds potential energy, U, at x=A, so once  the speed has decreased back to 0, the potential energy at its new point must be equidistant from the equivalent for all the energy to have been retransfered to from kinetic to potential. Therefore, the new position must be of the same magnitude but opposite phase of the original: -A.
Once released from this new position, the block will move in the same manner as when it was first released. It moves towards the equilibrium position, surpasses it, and ends up back at x=A with the potential energy restored.
Frequency and period are inversely proportional to each other. Therefore, when the period increases by a factor of two, the frequency decreases by the same factor (ie. it is divided by 2, or halved).

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Additional links

Alesa Rabson & Maria Forero - Simple Harmonic Motion