# Find the time when the spring reaches its max height

• Helly123

## Homework Equations

f = Spring's constant * ##\Delta x##
Potential Energy = ##\frac{1}{2} kx^2##

## The Attempt at a Solution

I try to find constant of the springs
Fabove + Fbelow - W = 0
k*(0,2 m) + k*(0,2m) - ##\frac{1}{2}## * 9,8 = 0
0,4k = 4,9
k = 12,25 N/m

can I use Conservation energy to find v when small object is on the bottom, like this?
Ep when rest + Ek when rest = Ep bottom + Ek bottom
##\frac{1}{2} k x^2## + 0 = 0 + ##\frac{1}{2} m v^2##
##\frac{1}{2} *12,25 * {0,3}^2## = ##\frac{1}{2} \frac{1}{2} v^2##
v^2 = 12,25 * 0,09 * 2
I am not sure what to do next. Can you help me?

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It can be shown that the mass will execute regular harmonic motion about its new point of equilibrium in Figure 2. Recast the question to "what fraction of the period T is needed for the mass to reach the maximum height" and you will have your answer.

On Edit: Of course you will have to find a value for the period.

It can be shown that the mass will execute regular harmonic motion about its new point of equilibrium in Figure 2. Recast the question to "what fraction of the period T is needed for the mass to reach the maximum height" and you will have your answer.

On Edit: Of course you will have to find a value for the period.

Hint: Use this equation, but spring constant "k" here is not as simple ;)

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Helly123
Weight force = k*x
K system = mg/x
20 cm for 2 springs
Displacement of one spring is 10 cm
K system = 49
I add k = 49 to find T
T = 0.63 sec
Until reach max height = (1/2)T = 0.315 sec
Its wrong.
What should i do?

What should i do?
Displacement of one spring is 10 cm
Why do you say that? Look at the picture you posted (Figure 2). Does it look like either spring is displaced by 10 cm? If you move the mass back up to the dotted line, by how much does each spring move?

Why do you say that? Look at the picture you posted (Figure 2). Does it look like either spring is displaced by 10 cm? If you move the mass back up to the dotted line, by how much does each spring move?
It moves by 20 cm.. i get wrong k. Didnt i? So, can you give me clue to find k? When the springs pressed downward, the forces applied to the mass will be
-F downward + F spring 1 + F spring 2 - W = m*a
Is it right?

It moves 20 cm.. i get wrong k. Didnt i? So, can you give me clue to find k? Can i get the max height by using Height = vo*t -##\frac{1}{2}at^2##

For now, focus on finding k with extension of the spring.
Since each spring moves 20cm, you can interpret the new k as double of the original k if you visualise it as one spring moving 20cm instead of 2 springs moving 20cm each.

For now, focus on finding k with extension of the spring.
Since each spring moves 20cm, you can interpret the new k as double of the original k if you visualise it as one spring moving 20cm instead of 2 springs moving 20cm each.
Yes. Do you mean
2k*##\Delta x## = mg
K = 12,25 N/m

Yes. Do you mean
2k*##\Delta x## = mg
K = 12,25 N/m

yep. You should be able to move on from here.

yep. You should be able to move on from here.
I get T = 1.27 sec
What should i do next? In this quiz not allowed calculator.. how can i do calculation without calculator?

I get T = 1.27 sec
What should i do next? In this quiz not allowed calculator.. how can i do calculation without calculator?

To complete it, refer to post #2

Oh, btw i just realized, K is 12.50N/m, not 12.25. (edit: oops missed out the decimal points)

using that, your square root can be evaluated by observation.

You should be able to get the answer when you approximate pi to be 3 :)

To complete it, refer to post #2

Oh, btw i just realized, K is 12.50N/m, not 12.25. (edit: oops missed out the decimal points)

using that, your square root can be evaluated by observation.

You should be able to get the answer when you approximate pi to be 3 :)
that cannot be.. it is exactly 12.25 that i get..

that cannot be.. it is exactly 12.25 that i get..

Oh, I forgot that I used 10ms^-2 for g instead of 9.8

I avoided the 9.8 as it couldn't give me a round number to evaluate the square root by observation (without calc)

If this quiz permitted calculator, go ahead and use the 9.8. I have a feeling this question was not designed with the absence of calculator in mind.

To complete it, refer to post #2

Oh, btw i just realized, K is 12.50N/m, not 12.25. (edit: oops missed out the decimal points)

using that, your square root can be evaluated by observation.

You should be able to get the answer when you approximate pi to be 3 :)
Why the pi is 3. And what fraction of T should i approach to get max high?

Why the pi is 3. And what fraction of T should i approach to get max high?

You mentioned that you arent allowed to use calculators, so I just took the simplest round numbers of all the values to get a rough approximation of the answer, which seems to be sufficient for this question.

You got the fraction of T right in post #4.

You mentioned that you arent allowed to use calculators, so I just took the simplest round numbers of all the values to get a rough approximation of the answer, which seems to be sufficient for this question.

You got the fraction of T right in post #4.
How is the process of getting pi equals to 3?

How is the process of getting pi equals to 3?

for no-calculator calculations, just round off pi = 3 to get easy numbers to do mental sums with. No specific process. You can try with 3.14, but have fun with that lol

for no-calculator calculations, just round off pi = 3 to get easy numbers to do mental sums with. No specific process. You can try with 3.14, but have fun with that lol

Well yes i got 0.6.
Hmm okay if the answer is 0.45... I have a feeling the k they used is the small k instead of the big one you found... you need to check with your tutor about that. Was taught to use big K in school anyways.

Well yes i got 0.6.
Hmm okay if the answer is 0.45... I have a feeling the k they used is the small k instead of the big one you found... you need to check with your tutor about that. Was taught to use big K in school anyways.
I do not have any tutor lol. I study by myself. Small k how is it?

I do not have any tutor lol. I study by myself. Small k how is it?

If we use small k, based on my approximations it is around 0.42-0.45 (depends on whether you want to use calculator or not)

This is kinda weird to me too :/

Edit: do you have the source of the original question?

Edit 2: According to your image, the answer is option 4, which is indeed 0.60s

Forget the big and small ##k##. Consider the two springs as a single spring of equivalent constant ##k##. At the equilibrium position, ##kd=mg~\rightarrow \frac{m}{k} = \frac{d}{g}##. The period then is $$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{d}{g}}$$
Plug into find ##T## then divide by two. Use a calculator.

Helly123
Well yes i got 0.6.
Hmm okay if the answer is 0.45... I have a feeling the k they used is the small k instead of the big one you found... you need to check with your tutor about that. Was taught to use big K in school anyways.
I do not see the need to find the constant for the individual springs. The pair have some effective constant, k. The ratio of this to the mass is immediately derivable from the given equilibrium point, and leads quickly to π/7 ≈ 0.45 s.

Edit: pipped by Kuruman.

Helly123
I do not see the need to find the constant for the individual springs. The pair have some effective constant, k. The ratio of this to the mass is immediately derivable from the given equilibrium point, and leads quickly to π/7 ≈ 0.45 s.

Edit: pipped by Kuruman.

Sorry, I don't get the pi/7 step: a brief explanation would be appreciated

Oh yea I realized I should've used "effective k" instead of big and small, apologies for the confusion.

Sorry, I don't get the pi/7 step: a brief explanation would be appreciated
Plug d=0.2 and g=9.8 into the equation for T/2 derived from that in #22.

Plug d=0.2 and g=9.8 into the equation for T/2 derived from that in #22.

Oh i see.. oh LOL i forgot to x2 for my k that I found... no wonder there was some confusion <<sighs>>

Oh... why its so easy for you. Lol. I am frustrated because of it. Thanks everyone!

I do not see the need to find the constant for the individual springs. The pair have some effective constant, k. The ratio of this to the mass is immediately derivable from the given equilibrium point, and leads quickly to π/7 ≈ 0.45 s.

Edit: pipped by Kuruman.
Forget the big and small ##k##. Consider the two springs as a single spring of equivalent constant ##k##. At the equilibrium position, ##kd=mg~\rightarrow \frac{m}{k} = \frac{d}{g}##. The period then is $$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{d}{g}}$$
Plug into find ##T## then divide by two. Use a calculator.

But. How to find k? And use it as m/k?

But. How to find k? And use it as m/k?
You don't need either of m or k individually. All you need for the period is the ratio between the two, and this can be calculated directly from the equilibrium information: mg = kh, where h is the spring displacement that resulted in equilibrium.

Helly123
You don't need either of m or k individually. All you need for the period is the ratio between the two, and this can be calculated directly from the equilibrium information: mg = kh, where h is the spring displacement that resulted in equilibrium.
Yes.. so. For this kind of quiz (in case i bump into one again) i don't need m or k. Right?

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.

If you have the ratio that's all you need.
Ok thanks. :)

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.

Helly123
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.

Helly123
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.