Spring from a ceiling problem

[RJLiberator]

Homework Statement

A massless spring with no mass attached to it hangs from the ceiling. Its length is 0.2
meter. A mass M is now hung on the lower end of the spring. Imagine supporting the mass
with your hand so that the spring remains relaxed (that is, not expanded or compressed),
then suddenly remove your supporting hand. The mass and spring oscillate. The lowest
position of the mass during the oscillations is 0.1 m below the place it was resting when
you supported it.
(a) What is the frequency of oscillation?
(b) What is the velocity when the mass is 0.05 m below its original resting place?

Homework Equations

t = sqrt(2*d/g)
g = 9.8
frequency = w/2pi
w^2=g/l
x=Acos(wt+Φ)

The Attempt at a Solution

These problems always throw me off. I do not want to be thrown off this time around! Please check my work. I am somewhat confident in it.

1. To find frequency of oscillating we find w. w is easy enough by the equation w^2=g/l. Since the length is 0.2 and g = 9.8 this comes out to 7.
question: I used the correct length correct? Can I use this equation? This was giving to me in my pendulum notes but we are working with a spring... Is this incorrect?

2. With w, it is easy to see frequency = w/2pi, so 7/2pi is thus the frequency and part a.

3. To find velocity we need the position function which is x = 0.1sin(7*t)
or equally 0.1cos(7t+pi/2)

4. Take the derivative which becomes velocity = 0.7cos(7t)

5. Find t for 0.05m. t=sqrt(2*0.05/9.8) = 0.101

6. Plug t into velocity equation to get 0.7cos(7*0.101) = 0.6999 = 0.7m/s

Did I use wrong equations and get the wrong answer? I am somewhat skeptical about my final velocity as it is saying that cos(0)=1 and so it takes on 0.7.

 

[Dr. Courtney]

You need to first use the given info and compute the spring constant. Then use the spring constant and the mass to solve the rest of the problem.

Don't skip to the solution to the equation of motion. Use Newton's second law to express the equation of motion.

This will allow computing the frequency in terms of mass and spring constant.

A picture never hurts either.

 

[RJLiberator]

Then use the spring constant and the mass to solve the rest of the problem.

We do not have the mass, otherwise I would have done that for sure. Is there something I am missing here?

 

[haruspex]

We do not have the mass, otherwise I would have done that for sure. Is there something I am missing here?

A bit of dimensional analysis helps here. You know g, and you are given some distances, so all of the information you have involves distances and times only. That may well be enough to find a frequency and a velocity, since those also involve only distance and time. Knowing the mass cannot help unless you also know something else that involves a mass dimension, such as the spring constant. So there is a very good chance that you do not need to know either.
Proceed by letting the mass be M and deriving equations involving M. Expect M to disappear later.

 

[TSny]

1. To find frequency of oscillating we find w. w is easy enough by the equation w^2=g/l. Since the length is 0.2 and g = 9.8 this comes out to 7.
question: I used the correct length correct? Can I use this equation? This was giving to me in my pendulum notes but we are working with a spring... Is this incorrect?

The system is not oscillating as a pendulum. So you are using the wrong equation for the frequency.

You will not be able to determine the mass or the spring constant. However, you should be able to determine the ratio k/m, which is all you need to answer the questions.

 

[Dr. Courtney]

Proceed by letting the mass be M and deriving equations involving M. Expect M to disappear later.

This plan nearly always succeeds in intro physics when you do not have the mass and think you need it.

 

[RJLiberator]

The system is not oscillating as a pendulum. So you are using the wrong equation for the frequency.

You will not be able to determine the mass or the spring constant. However, you should be able to determine the ratio k/m, which is all you need to answer the questions.

Boom. Been staring at this problem for the past few days, I think I finally have it.

Hooke's Law: F = -kx
k = F/x when looking at just magnitudes
since F = m*g
k = mg/x

now, k/m = g/x = w^2
g = 9.81 m/s^2
x = 0.2m (Am I using the right measure for x here? 0.2m length is the resting length of the string before mass was added, from what I see, that is the number we are supposed to use. Or am I to use 0.2m + 0.1m = 0.3m as the stretched out length??)

so w^2 = 9.81/0.2
w = sqrt(9.81/0.2) = 7.00 s^-1
this agrees with my first calculation of w.

Did I do something wrong here or is this the right approach?

 

[TSny]

Hooke's Law: F = -kx
k = F/x when looking at just magnitudes
since F = m*g
k = mg/x

I'm not sure of what you did here. Why did you say that the gravitational force is equal to the spring force? Is this always true or is it true for only one particular value of x?

To make sure that you have the correct interpretation of x in the formula F = -kx, can you state the value of x at the initial position where the block was released?

What is the value of x when the block gets to its lowest point?

 

[RJLiberator]

I'm not sure of what you did here. Why did you say that the gravitational force is equal to the spring force? Is this always true or is it true for only one particular value of x?

We have some mass, M hanging from a massless spring from the ceiling. The massless spring is initially 0.2 m long prior to attaching the mass. When the mass is attached and let to reach it's amplitude, it falls another 0.1 m down from the ceiling. This is a total of 0.3 m from the ceiling.

I am using Hooke's law where k = F/x. Force in this problem is equal to m*a = m*g since gravity is the acceleration / only force acting on the mass here.
Now, using the hints in this thread, I saw that k/m = g/x = w^2.

To make sure that you have the correct interpretation of x in the formula F = -kx, can you state the value of x at the initial position where the block was released?

The value of x at the initial position was 0.2 m long. The resting point for the spring is 0.2 m. That's why I think that value is the x value.

What is the value of x when the block gets to it's lowest point?

If we let the equilibrium position be x = 0, then the amplitude is 0.1m and the lowest point is x = 0.1m or -0.1m depending on the orientation of your plot.

:) ?

 

[haruspex]

Hooke's Law: F = -kx
k = F/x when looking at just magnitudes
since F = m*g
k = mg/x

You need to be clearer in your mind (and preferably in your postings too) regarding what your variables represent.
The F in Hooke's law refers to the tension. The F in F=mg is the gravitational force. Under what circumstances will these be equal? What does that imply for the meaning of x in the last equation above?

 

[RJLiberator]

You need to be clearer in your mind (and preferably in your postings too) regarding what your variables represent.
The F in Hooke's law refers to the tension. The F in F=mg is the gravitational force. Under what circumstances will these be equal? What does that imply for the meaning of x in the last equation above?

I see.
That is a subtle problem.

So, in actuality, the x value should be the change in x. The change in x is 0.1m and NOT 0.2m which I had originally thought was the case.

 

[haruspex]

So, in actuality, the x value should be the change in x. The change in x is 0.1m and NOT 0.2m which I had originally thought was the case.

Closer, but still not right. Please try to answer the question I posed in the previous post: under what circumstances will the two forces be equal?

 

[RJLiberator]

The F in Hooke's law refers to the tension. The F in F=mg is the gravitational force. Under what circumstances will these be equal?

I think I have it:
It is when we are at equilibrium. Here the tension is equal to the force applied when it is in equilibrium
And therefore, 0.2m was actually the correct x value as we are describing the system at equilibrium and using the equations giving by it.

 

[haruspex]

I think I have it:
It is when we are at equilibrium. Here the tension is equal to the force applied when it is in equilibrium
And therefore, 0.2m was actually the correct x value as we are describing the system at equilibrium and using the equations giving by it.

Yes, the x you calculated is the equilibrium extension with the mass suspended. But that is not 0.2m. We do not know what it is yet.
You have:
Relaxed length = initial length before release = 0.2m.
Maximum extension = 0.1m.
Equilibrium extension = x.
What is the relationship between the three extensions?

 

[RJLiberator]

Oh man, classical mechanics is the death of me!
Such a joy, but wow, how many times can I be wrong. ^_^.

So what is happening is that when the mass is attached, it actually reaches the lowest point of 0.3m spring length.
So this is the number I should be using.
This makes sense, because that is where the system is at rest, 0.3m stretched spring where the forces cancel each other out.

 

[haruspex]

So what is happening is that when the mass is attached, it actually reaches the lowest point of 0.3m spring length.
So this is the number I should be using.
This makes sense, because that is where the system is at rest, 0.3m stretched spring where the forces cancel each other out.

No, 0.1m is the maximum extension after the system is released from the relaxed length.
If the system oscillates without loss, what is the minimum extension in each cycle?
Given the minimum and maximum extension each cycle, what is the equilibrium extension?

 

[RJLiberator]

If the system oscillates without loss, what is the minimum extension in each cycle?

Well, 0.1 is the amplitude, so 0.1+0.1 = 0.2 as the oscillation cycle displaced. (total)

So the minimum extension should also be 0.1 since the system goes on without loss, correct?

The equilibrium extension is 0.1m away from equilibrium.

How do we get a different number than 0.1m as amplitude or extension if nothing is lost and this is SHM?

 

[haruspex]

Well, 0.1 is the amplitude,

what is the minimum extensions each cycle?
What is the amximum extension each cycle?
So what is the amplitude?

 

[RJLiberator]

You pose a question that I am having trouble visualizing, let me explain.

Maximum extension: 0.1m
This is brought on by the fact that it reaches this distance as its low point per the question statement.
Minimum extension: 0.1m or 0m.
Here, perhaps I am not understanding the question. How is there a minimum extension each cycle? If there is no loss here, then the oscillation will be simple harmonic motion and continue on going for 0.1m below the initial position to 0.1m above the initial position, continuing. This would imply that hte minimum extension is = maximum extension = 0.1m. Perhaps you mean that the minimum is when it was at rest at 0.2m ?
Amplitude: 0.1m

If we look at a diagram, we see the start = 0.2m, the low point = 0.3m the return heigh point = 0.1 m. Is this what you mean?

 

[haruspex]

to 0.1m above the initial position, continuing.

How will it go above the release position? Think about energy.

 

[RJLiberator]

Hm, have I made a mistake? Oh my!

So when we consider Energy, at the release point, the energy must all be potential as it is at rest.
So we would then be saying that once it reaches the return point, it then has to go back down, so we'd say that :

At 0.025m is the equilibrium now.
0.03m is the lowest point
0.02m is the highest point.

Therefore, the amplitude is actually 0.05m

The max extension is 0.1 m from the original 0.2m mark. The minimum extension is 0.0m when it returns to that initial point.

 

[haruspex]

Hm, have I made a mistake? Oh my!

So when we consider Energy, at the release point, the energy must all be potential as it is at rest.
So we would then be saying that once it reaches the return point, it then has to go back down, so we'd say that :

At 0.025m is the equilibrium now.
0.03m is the lowest point
0.02m is the highest point.

Therefore, the amplitude is actually 0.05m

The max extension is 0.1 m from the original 0.2m mark. The minimum extension is 0.0m when it returns to that initial point.

Yes, except that you appear to have inserted an extra 0 in some of those, making the distances a tenth of what they should be.

 

[RJLiberator]

Absolutely.
0.2 is the start
0.3 is the low
0.25 is the new equilibrium
0.05 is the amplitude.

So now, k/m = g/x, where x is 0.05 as it is the change in x?
"Let X be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched)." via wikipedia. So we can read this as the spring is being stretched 0.05m from its relaxed position.

 

[haruspex]

Absolutely.
0.2 is the start
0.3 is the low
0.25 is the new equilibrium
0.05 is the amplitude.

So now, k/m = g/x, where x is 0.05 as it is the change in x?
"Let X be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched)." via wikipedia. So we can read this as the spring is being stretched 0.05m from its relaxed position.

Yes.

 

[RJLiberator]

Marvellous! You have gone beyond great lengths to help me with this problem. I must have had poor thinking over 10 times on this issue.

Now, with part b) what is the velocity when the mass is at 0.05m from it's original position, well, since this is the new equilibrium point, we know that the velocity here is maximized.
It is A*w which I find to be 0.05m*14.01rads/sec = 0.700m/s

 

[haruspex]

Marvellous! You have gone beyond great lengths to help me with this problem. I must have had poor thinking over 10 times on this issue.

Now, with part b) what is the velocity when the mass is at 0.05m from it's original position, well, since this is the new equilibrium point, we know that the velocity here is maximized.
It is A*w which I find to be 0.05m*14.01rads/sec = 0.700m/s

Not sure where you are going wrong but that cannot be the right answer. As a check, consider the speed the mass would have if dropped 0.05m with no spring attached.
The easiest route might be to draw yourself a matrix, 3 columns for energy - the two sorts of PE and the KE - and three rows for three positions of the mass: top, middle and bottom. Fill in the nine entries in terms of the symbols m, g, A, k, v_max. Energy conservation gives you two equations.

 

[RJLiberator]

I don't understand tho, how could I have made an error?

I calculated w by w^2 = g/x => sqrt(9.81/0.05) = 14.01
We find the amplitude, A, to be equal to 0.05m.

The max velocity is at 0.05m since that is the equilibrium position. A*w = 14.01*0.05 = 0.700m/s.

 

[TSny]

I think 0.70 m/s is correct. It is a bit surprising how fast it's moving after just 5 cm of "fall".

 

[haruspex]

I think 0.70 m/s is correct. It is a bit surprising how fast it's moving after just 5 cm of "fall".

Yes, sorry - my mistake.

 

[RJLiberator]

Scared the beejeebus out of me after how much time I've spent on this problem in the past few days :D.

Thank you guys for the help, as always. Haruspex, you are simply amazing :D.

 

[RJLiberator]

There is a part c and part d of this problem.
I'm almost at my limit, but I felt like posting this here if you guys wanted to review my work.
If you want, you can just reply "Wrong or Right," and I'd be happy. If not, no bothers, I didn't feel like this deserved it's own thread.

A second mass of 0.3 kg is added to the first mass, making a total of M+0.3kg. When this system oscillates, it has half the frequency of the system with mass M alone. (c) What is the value of M? (d) Where is the new equilibrium position?

c) I use a relationship
w^2 = k/M from original problem in this thread
and (1/2*w)^2 = k/(M+0.3)

With a little algebraic manipulation we see 3M = 0.3 so M = 0.1 kg is the answer.

d) I use w^2 = k/M from original problem and our new mass M = 0.1 to find spring constant K.
K = 19.628.
Now, I use this information and apply it to the equation to find amplitude K = mg/x where m is now 0.3+0.1 = 0.4 and x is what we are looking for.
Solving this gives me x = 0.2 from the initial point.

Using my knowledge I learned in this thread, I know this is 0.4m from the ceiling and so the new equilibrium position is now 0.3m.I feel confident with these answers and my strategy to tackle them, but as this thread has shown, my understanding of classical mechanics is tarnished :).
Are my strategies correct?

 

[haruspex]

I agree with your answers, except that I don't understand the final conversion to .3m in d. Where are you measuring it from to get that?

You could get d a little more easily. If the equilibrium extension for the original mass was .05m, the eq. ext. for four times the mass must be four times that.

 

[RJLiberator]

I get .3m from looking at the initial experiment.

The initial parts a and b saw that when the mass went 0.1 meters beneath the 0.2m initial starting point, that the equilibrium point was 1/2 of 0.1 m at 0.05m lower than 0.2m.
So I applied that same practice when the mass dropped the string 0.2 meters more to 0.4meters. So the oscillation is from 0.4 meter mark to 0.2 meter mark and therefore the 0.3 mark must be the equilibrium :).

I'm glad that I am starting to get this problem. It's clear to me that I've learned (something) in this thread.

 

[haruspex]

So I applied that same practice when the mass dropped the string 0.2 meters more to 0.4meters. So the oscillation is from 0.4 meter mark to 0.2 meter mark and therefore the 0.3 mark must be the equilibrium :).

But the .2m you calculated was from zero extension to the new equilibrium (4x0.05), not to the new lowest point. The new lowest point would be .4m from zero extension, making it a total length of .6m.