Oscillation of a vertical spring

In summary: The top end's motion is not just its acceleration. It is also its displacement. And its acceleration is not just the acceleration of its acceleration. It also includes the acceleration of its displacement.z is the top end's position. Then x(t) = z(t) -...Ok, I think I see what you mean. So then I have ##\ddot{L} = \ddot{x} - g = \ddot{z} - g - g##, where I am assuming the z is the displacement of the top end from its equilibrium position and the double g is the two terms that come from the acceleration of the top end's displacement. Then the equation is ##m\ddot{L
  • #1
CAF123
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Homework Statement


A mass m hangs in equilibrium at the lower end of a vertical spring of natural length a, extending the spring to be a length b.

1) Show that the frequency for small oscillations about the point of equilibrium is ##\omega = \sqrt{g/(b-a)}##

2) The top end of the spring is made to oscillate vertically with a displacement ##c\sin (nt)##, moving downwards at t=0. Show that as a result, the length of the spring L(t) obeys: $$\ddot{L} + \omega^2 L = \omega^2 b + cn^2 \sin (nt)$$

The Attempt at a Solution



1)Write the eqn of motion: ##m\ddot{y} = -mg + k(y - a)##. In equilibrium, y = b, so ##mg = k(b-a) \Rightarrow k = mg/(b-a)##. Taking the period of a mass on a spring in the case of small oscillations, ##T = 2 \pi \sqrt{m/k}##, I get that ##T = 2 \pi \sqrt{(b-a)/g} \Rightarrow \omega = 2 \pi/T = 2 \pi \cdot \sqrt{g/(b-a)} \frac{1}{2 \pi}## which gives the result. Is this ok - am I allowed to simply quote the formula for the period of a mass on a spring? (I assumed yes, since it is given in the question that we are interested in small oscillations, so the formula is valid).

2) If it is made to oscillate vertically with that displacement then the spring must be under some sort of external driving force. I haven't been able to make much progress with this one. I know the displacement is given by y = y(t) = c sin (nt). This means the external driving force causing this is of the form ##mcn^2 sin(nt)##. So ##m\ddot{L} = mg - k(L - c \sin(nt)) - mcn^2 \sin(nt)##, but this does not resemble what I need to show.

Many thanks.
 
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  • #2
Do not assume any force on the top end. Denote its location with some variable, say z. Denote the location of the other end with x. What forces act on it? What is its acceleration?
 
  • #3
voko said:
Do not assume any force on the top end. Denote its location with some variable, say z. Denote the location of the other end with x. What forces act on it? What is its acceleration?

Is the spring attached to a wall at the top end? What is the 'it' you are referring to? The only forces that act on the spring is the external force (that makes it undergo that displacement), and a restoring force. (not gravity on the spring since it is not assigned a mass).
 
  • #4
The top end is attached to some mechanism that makes it oscillate as indicated. Denote the location of the top end as z. Do not analyze what force is required to make it move that way, just assume it does move that way. The other end of the spring has point mass m attached to it. Denote its location as x. Analyze the forces acting on the mass.
 
  • #5
voko said:
The top end is attached to some mechanism that makes it oscillate as indicated. Denote the location of the top end as z. Do not analyze what force is required to make it move that way, just assume it does move that way. The other end of the spring has point mass m attached to it. Denote its location as x. Analyze the forces acting on the mass.

Ok, so the mechanism causes the mass to be displaced by an amount ##c \sin (nt)##. We are given that the natural length of the spring is a and it has some length L. So then writing the forces on the mass: ## m\ddot{L} = k(L - c\sin (nt)) -mg##
 
  • #6
That is close, but not quite. Specifically, why do attach the mass to the second derivative of the length? It is the mass whose acceleration your trying to find here. I personally find it easier to get the basic equation in terms of the two ends' positions first, then convert to L(t).
 
  • #7
voko said:
That is close, but not quite. Specifically, why do attach the mass to the second derivative of the length? It is the mass whose acceleration your trying to find here. I personally find it easier to get the basic equation in terms of the two ends' positions first, then convert to L(t).

Actually, I think what I have written makes no sense at all since the mass of the spring is negligible. If I denote the position of the mass as x, then ##m\ddot{x} = k(L - c\sin(nt)) -mg##. I don't think we are interested in x, it is L that we want. What would be my next step?
 
  • #8
CAF123 said:
##k(L - c\sin(nt))##.

This term is wrong. What you have in brackets must be extension from the spring's natural length a.

I don't think we are interested in x, it is L that we want. What would be my next step?

Once you get the rhs right, the next step would be to express the lhs in terms of L and the top end's motion.
 
  • #9
voko said:
This term is wrong. What you have in brackets must be extension from the spring's natural length a.

Ok, so I should have ##\ddot{x} = \frac{k}{m}(a - c \sin(nt)) - g##

Once you get the rhs right, the next step would be to express the lhs in terms of L and the top end's motion.

Is the acceleration of the mass only slightly greater than the acceleration of the spring? So ##\ddot{L} = \ddot{x} - g##? For the top end motion, I am not really sure how to analyze here. The acceleration (I think) is ##cn^2 \sin(nt)##
 
  • #10
CAF123 said:
## \frac{k}{m}(a - c \sin(nt)) ##

How come this is independent of the bottom end's motion?

Is the acceleration of the mass only slightly greater than the acceleration of the spring? So ##\ddot{L} = \ddot{x} - g##? For the top end motion, I am not really sure how to analyze here. The acceleration (I think) is ##cn^2 \sin(nt)##

z is the top end's position. Then x(t) = z(t) - L(t).
 
  • #11
voko said:
How come this is independent of the bottom end's motion?
In equilibrium, x = b without that driving force, so I think my expression should become: $$\ddot{x} = \frac{k}{m} (x - a - c\sin (nt)) - g$$


z is the top end's position. Then x(t) = z(t) - L(t).

I know that x represents the position of the mass, L(t) the length of the spring but is z simply the position of the end of the spring?

So then, $$\ddot{z} - \ddot{L} = \frac{k}{m}(x - a - c\sin(nt)) - g, $$ where ##\ddot{z} = -cn^2 \sin(nt)##
 
  • #12
Note the description says "moving downward at time t = 0". So is z = c sin nt or z = - c sin nt?
 
  • #13
z would be -c sin(nt). So ##\ddot{z} = cn^2 \sin(nt) ##.

My eqn is then:$$\ddot{z} - \ddot{L} = \frac{k}{m}( - L - a ) - g$$and then I should just reexpress in terms of ##\omega##?
 
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  • #14
Again, the term in the brackets must be the extension from length a.
 
  • #15
I have the eqn in the form: $$\ddot{L} - \omega^2L = \omega^2 b + cn^2 \sin (nt),$$ but I appear to have a minus error in the second term on the LHS.

Using the form given in the question, I proceeded to find L. The general soln to the homogeneous eqn is ##L_H(t) = A\cos(\omega t) + B \sin (\omega t). ## I am thinking to find the particular solution that I should find the soln to the two eqns ##\ddot{L} + \omega^2 L = \omega^2 b## and ##\ddot{L} + \omega^2 L = cn^2 \sin(nt)## separately and then make a linear combination of the two solns.

However, in order to find the two parameters, A and B, I require two initial conditions. I think one is L(0) = -x and the other ##\dot{L}(0) = -nc?##
Thanks.
 
  • #16
voko said:
Again, the term in the brackets must be the extension from length a.

All I did was replace x by z - l, z = c sin(nt) into the eqn at the bottom of post 11. Why is my eqn in post 11 wrong?
 
  • #17
Re the sign error, see #14.

I am not sure what initial conditions you are supposed to have. If the assumption is that the point mass is in equilibrium at t = 0, then you conditions seem correct.
 
  • #18
CAF123 said:
All I did was replace x by z - l, z = c sin(nt) into the eqn at the bottom of post 11. Why is my eqn in post 11 wrong?

Assume there is no sine term in the equation. Then it should be the usual harmonic oscillator equation. Is it the case?
 
  • #19
voko said:
Assume there is no sine term in the equation. Then it should be the usual harmonic oscillator equation. Is it the case?

If there is no sin term, then I have -k(x-a) on the RHS. Since x>a, then this represents an extension, so it is in the form of Hookes Law, no?
 
  • #20
x > a is incorrect, because x < 0 and a > 0.
 
  • #21
voko said:
x > a is incorrect, because x < 0 and a > 0.
I see. So I should have -k(-x-a) which in my case would give -k(-c sin(nt) -x - a). This is true since we have motion downwards and so the spring is stretched in the -ve axis?
 
  • #22
That's too many minus signs :)

The natural location of the spring's bottom end is at -a. When it is at x, the extension is X = x - (-a) = x + a. Finally, the restoring force is -kX = -k(x + a). In terms of L, it can be seen directly that the extension X = -(L - a), so the force is -kX = k(L - a).
 
  • #23
I think it is the minus signs that I am getting confused with:
voko said:
The natural location of the spring's bottom end is at -a. When it is at x, the extension is X = x - (-a) = x + a.
So here x < a (i.e more negative than a)

In terms of L, it can be seen directly that the extension X = -(L - a), so the force is -kX = k(L - a).
But if z = -c sin(nt) then this would give -2c sin(nt)?

If the spring is at a position -a, and it is then displaced in the downwards direction by some displacement -z. So X = -z-(-a) = -z + a = x - l + a. What is wrong with this analysis?
 
  • #24
CAF123 said:
I think it is the minus signs that I am getting confused with:

So here x < a (i.e more negative than a)

That is always the case because x is negative, and a is positive. a is the length, and x is the position.

But if z = -c sin(nt) then this would give -2c sin(nt)?

If the spring is at a position -a, and it is then displaced in the downwards direction by some displacement -z. So X = -z-(-a) = -z + a = x - l + a. What is wrong with this analysis?

If both ends are moving, then you pretty much have to deal with L = z - x. The natural value of L is a. Then extension X = -(L - a). You might be wondering why there is the minus sign in front of (L - a). This is because we want the extension of the elongated string to be negative. And why is that? Because that's how we have it when the top end is fixed.
 
  • #25
voko said:
That is always the case because x is negative, and a is positive. a is the length, and x is the position.



If both ends are moving, then you pretty much have to deal with L = z - x. The natural value of L is a. Then extension X = -(L - a). You might be wondering why there is the minus sign in front of (L - a). This is because we want the extension of the elongated string to be negative. And why is that? Because that's how we have it when the top end is fixed.

I am wondering how this conforms with what I wrote to obtain part 1) of this question. There I said the extension was y - a, so I get a spring force -k(y-a). In equilibrium, I then set y = b. However, from the above discussion, this should be -y-a and then I set -y = b, right?
 
  • #26
It is completely arbitrary what direction we label positive. It usually makes sense to label the upward direction positive; then any negative extension - pulling the spring down - results in an in a positive (upward) restoring force.

But you could equally label the downward direction positive. Then any "pulldown extension" is positive. This is especially true when you deal with statics, where things are usually more intuitive. So there is no mistake in the way you did part 1). Of course, you could have said that the extension was x - (-a) = x + a, thus getting (-b) + a = a - b as the extension at the equilibrium, and then having k(b - a) as the restoring force. In the end, you did not really care about the sign of the restoring force, you knew it would be opposite of the force of gravity, what you really cared about was its absolute value. In kinematics and dynamics, however, one typically has to be a bit more disciplined about directions and sign conventions.
 
  • #27
voko said:
It is completely arbitrary what direction we label positive. It usually makes sense to label the upward direction positive; then any negative extension - pulling the spring down - results in an in a positive (upward) restoring force.

But you could equally label the downward direction positive. Then any "pulldown extension" is positive. This is especially true when you deal with statics, where things are usually more intuitive. So there is no mistake in the way you did part 1). Of course, you could have said that the extension was x - (-a) = x + a, thus getting (-b) + a = a - b as the extension at the equilibrium, and then having k(b - a) as the restoring force. In the end, you did not really care about the sign of the restoring force, you knew it would be opposite of the force of gravity, what you really cared about was its absolute value. In kinematics and dynamics, however, one typically has to be a bit more disciplined about directions and sign conventions.
Actually in 1) I also had downwards as negative.
 
  • #28
I would say it was a mix.

You had +k(y - a) on the RHS, while that should have been -k(y + a). And then you let y = b, while that should have been y = -b.

So you got +k(b - a), which is correct for "downward = negative".

And you had -mg, which is also correct in this convention.
 
  • #29
Ok now I am trying to recover the form of the expression in the question: Take up as positive:
Put the top end of the spring at the origin before it becomes displaced. The natural length of the spring is therefore -a. After the mechanism starts, the mass is displaced a distance -x. (-x < -a) So the spring force is -k(-x+a). I know that if the mass goes down by -x then the top end of the spring goes down by -x. So then l = z - x. The spring force in terms of l and z is then -k(l-z+a) = -k(l - csin(nt) + a). But that csin(nt) should have disapeared. What did I miss?
 
  • #30
The natural length is not -a. It is a and always positive - simply because it is a length. What you meant to say was the position of the bottom end at the natural length was -a, which is quite a mouthful.

The fallacy of your argument is this assumption: "I know that if the mass goes down by -x then the top end of the spring goes down by -x." The top and the bottom ends are obviously connected by the spring, but you can't assume one simply repeats the motion of the other. This is obvious when you consider the fixed top: the bottom can be at pretty much any position x, despite the fixed top's being, well, fixed.
 
  • #31
voko said:
The natural length is not -a. It is a and always positive - simply because it is a length. What you meant to say was the position of the bottom end at the natural length was -a, which is quite a mouthful.

The fallacy of your argument is this assumption: "I know that if the mass goes down by -x then the top end of the spring goes down by -x." The top and the bottom ends are obviously connected by the spring, but you can't assume one simply repeats the motion of the other. This is obvious when you consider the fixed top: the bottom can be at pretty much any position x, despite the fixed top's being, well, fixed.

The mass will undergo a greater displacement -c sin(nt). So I should have -k(-x - c sin(nt) +a) with l = z-x => -k(l-z -csin(nt) + a) => -k(l-(-csin(nt))-csin(nt) + a) = -k(l+a).
EDIT:No, this is still not right...but I cannot see where.
EDIT2: Actually, I think I have it. In my above setup, I should have -l = -x -(-z) = -x + z. So from -k(-x - csin(nt) + a), I have -k(-l-z - csin(nt) + a) = -k(-l - (-csin(nt)) - csin(nt) + a) = -k(-l + a), which I think is the same as what we got a few posts above in the discussion.
 
Last edited:
  • #32
You seem stuck on the idea that you can equate the motion of end end with that of the other. But that is not true. All you know is the motion of the top end, the motion of the bottom end is unknown. They are connected indirectly, via the force and acceleration.
 
  • #33
The bottom motion is connected to the top motion by expressing l in terms of z and x. (see my EDIT2 in my last post), thanks.
EDIT: I thought this fixed things but it messes the LHS up now. I think I was along the right lines in post 11, but I just can't seem to fix the negative signs.
 
Last edited:
  • #34
Okay let me try from scratch:

Consider the spring end to be at a distance -a. Now consider some displacement of the bottom end to -x. This means from the springs natural length, it has extended a distance
(-x +a). But there is also some displacement at the top end by -z. I believe these are related by the length of the spring: (-x + a) + z = l, so x = z + a -l.

Since the extension of the spring is (-x +a) this gives (l-z) (using x from above) and so the displacement of the bottom end is -k(l-z + a)

Writing the eqn of motion for the mass on the spring: ##m\ddot{x} = -k(l-z+a) -mg##. I haven't assumed anything about the displacement of the bottom mass here (only that z and x are related through l) but I still get a wrong sign and a stray z term. I don't see where I am going wrong.
 
  • #35
See picture. The ceiling is the zero of coordinate y and downward is the positive direction. The top of the spring is at x=csin(nt). The length of the spring is L. so y=x+L.

For he mass m, [itex]m\ddot {y}=mg-k(L-a)\rightarrow \ddot {x} +\ddot L =g-\frac{k}{m}(L-a)[/itex]

The relaxed length of the spring is a. When the hanged mass is rest, the length is b, so mg=k(b-a), as you have shown. The natural frequency is ω. ω2=√(k/m). So g=ω2(b-a).

Find the second derivative of x. Also substitute ω2 for k/m and ω2(b-a) for g.

ehild

Edit: The zero of y is not at the ceiling but at the point the upper end of the spring oscillates about.
 

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