Find the time when the spring reaches its max height

[haruspex]

Yes.. so. For this kind of quiz (in case i bump into one again) i don't need m or k. Right?

If you have the ratio that's all you need.

 

[Helly123]

If you have the ratio that's all you need.

Ok thanks. :)

 

[kuruman]

If you have the ratio that's all you need.

Even the ratio $m/k$ is not needed. All you need is the displacement of the springs to the new equilibrium position under the load and the acceleration of gravity. See post #22.

 

[haruspex]

Even the ratio $m/k$ is not needed. All you need is the displacement of the springs to the new equilibrium position under the load and the acceleration of gravity. See post #22.

Well, you do need m/k to use the equation for the period, but in this case m/k can be found from d/g.
It depends whether the question of what you need is purely in the context of questions like this one or more general.

 

[kuruman]

It depends whether the question of what you need is purely in the context of questions like this one or more general.

Yes, strictly in the context of the question. To the untrained eye, the equation $T=2\pi\sqrt{d/g}$ may seem to imply that when the oscillator is taken to the Moon, the period will decrease, just like the period of a pendulum.

 

[haruspex]

Yes, strictly in the context of the question. To the untrained eye, the equation $T=2\pi\sqrt{d/g}$ may seem to imply that when the oscillator is taken to the Moon, the period will decrease, just like the period of a pendulum.

Yes, good point. I find quite a lot of missteps in threads on this forum come from applying a memorised equation out of its rather narrow context. As a student, I had the advantage of not being able to remember reams of equations, so relied on the most fundamental ones.

 

[Helly123]

Yes, good point. I find quite a lot of missteps in threads on this forum come from applying a memorised equation out of its rather narrow context. As a student, I had the advantage of not being able to remember reams of equations, so relied on the most fundamental ones.

it's weird.. why there's g on periode and not depend on g..

 

[kuruman]

Because when $g$ decreases by some factor, $d = mg/k$ decreases by the same factor the ratio $d/g$ remains the same because it's equal to the ratio $m/k$ that remains the same.

 

[Helly123]

Because when $g$ decreases by some factor, $d = mg/k$ decreases by the same factor the ratio $d/g$ remains the same because it's equal to the ratio $m/k$ that remains the same.

haha.. Socrates ;)