An object of mass 0.109kg hangs from a long spiral spring. When pulled down 0.10m below its equilibrium position and released, it vibrates with a period of 2s .d. How much will the spring shorten if the object is removed?

Question

An object of mass 0.109kg   hangs from a long spiral spring. When pulled down 0.10m  below its equilibrium position and released, it vibrates with a period of 2s .d.	How much will the spring shorten if the object is removed?

anonymous · Answer

The period of oscillation doesn't depend on the length that it is extended, this is referred to as isochronus. We can use the equation for the time of oscillation which is given as$$T = 2 \pi \sqrt{ m \over k}$$We know all unknowns except $k$, which is what we want to solve for. After we determine $k$, we can use Hook's Law to determine the unloaded length of the spring. $$F = -kx$$where $F=-mg$. We can solve Hook's Law for $x$ which will be the change in length after the mass is removed.