# Distance traveled and period of a mass - spring - pulley system

songoku
Homework Statement:
A mass M is attached to a free pulley and the pulley is attached to wire S and elastic spring (see diagram). The spring will extend 2.0 mm per Newton tensile force. When M is stationary, the extension of spring is 98 mm. The mass of pulley, wire and spring are negligible. M is pushed slowly upwards until there is no spring restoring force, then released.
a. by how far will M move down before it starts going up again? (answer = 392 mm)
b. find the period of the motion (answer = 0.2 π second)
Relevant Equations:
Hooke's law

Simple harmonic motion
1.
How will the motion of M be? I assume wire S is inelastic so will M move downwards but not in straight line? (I mean M moves downwards but because the left side of pulley is connected to S, it will be static and the right side of pulley can go down along the extension of the spring so its movement will be curve). Or I assume wire S just like spool of thread coiling around the pulley so the pulley and M will move vertically downwards?

2.
Why the answer to question (a) is not 196 mm? When M is stationary and the spring extended by 98 mm, this will be equilibrium point of simple harmonic motion. When M is pushed slowly upwards until there is no spring restoring force, it means that the extension of spring is 0 so I think 98 mm will be the amplitude of simple harmonic motion. When M released, it will move downwards as far as twice the amplitude

Thanks Homework Helper
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I assume the pulley and mass move vertically. In the absence of measurements and distances, this seems the only sensible assumption. So I take the width of the pulley (diameter) to equal the distance between the anchor points for S and the spring.

I share your puzzlement. It looks as if the spring must stretch twice the movement of the mass.
They seem to say that the reverse is the case: the mass moves twice the distance of the spring stretch?

I do disagree with your 196 mm though.

• songoku and TSny
songoku
I share your puzzlement. It looks as if the spring must stretch twice the movement of the mass.
They seem to say that the reverse is the case: the mass moves twice the distance of the spring stretch?
Sorry I don't understand your hint. Does "the spring must stretch twice the movement of the mass" differ from "the mass moves twice the distance of the spring stretch"?

And I am also confused with the equation of motion:
a. When the spring extends by 98 mm and M is at rest
At this position, there are three forces acting on M: its weight, spring force, and tension S so Weight = kx + Ts

b. When there is no restoring force and M is released
At this position, there are only two forces acting on M: weight and tension S so Weight - Ts' = M.a

W - Ts' = M.a
kx +Ts - Ts' = M.a

I can not see this equation as simple harmonic motion equation because the value of Ts and Ts' are different and the acceleration is not directly proportional to displacement from equilibrium point. Or we assume Ts = Ts' (the tension of S is always constant)?

Thanks

Homework Helper
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It wasn't a hint! It was just saying that the person who set the question seems to be a bit confused. I assumed you had thought the same, because you already knew that their answer for (a) is wrong.

The point is, how much the spring stretches when the mass moves. Are they the same ?
Or to make it a multiple choice Q
a) mass moves 2 mm, spring stretches 2 mm
b) mass moves 2 mm, spring stretches 1 mm
c) mass moves 2 mm, spring stretches 4 mm
d) none of the above

The whole significance of this question lies in the role of the pulley.
A pulley can alter the amount of the movement and also the force.
For an ideal pulley these are in inverse proportion (due to conservation of energy: |work|or|force x distance| is the same on both sides of the pulley)

" ... a free pulley ... and elastic spring ... . The mass of pulley, wire and spring are negligible. "
To me that implies we have to assume the wire is light, inextensible and has no friction with the pulley or anything else. Therefore the tension is the same at all points along the wire.
Similarly the pulley has no effect other than that of an ideal pulley.

What you need to show is, that the force on the mass is proportional to the displacement of the mass from the equilibrium position. (obviously with the correct sense)

You need to be clear about what your variables stand for. Is x the displacement of the mass, or the extension of the spring? Or what? Where is the tension? In S, or in the string connecting the mass to the pulley?

You didn't say what Ts and Ts' are. I'd just stick to one T for tension in any one place, but realize that it is a variable.

Or we assume Ts = Ts' (the tension of S is always constant)?
No, the tension in the wire is not constant. It is attached to the spring, so for the spring to stretch tension must be applied by the wire

You need to consider just one position where the mass is not at equilibrium.
(The other position is just to allow you to calculate the mass.)

• songoku and Delta2
songoku
The point is, how much the spring stretches when the mass moves. Are they the same ?
Or to make it a multiple choice Q
a) mass moves 2 mm, spring stretches 2 mm
b) mass moves 2 mm, spring stretches 1 mm
c) mass moves 2 mm, spring stretches 4 mm
d) none of the above

The whole significance of this question lies in the role of the pulley.
A pulley can alter the amount of the movement and also the force.
For an ideal pulley these are in inverse proportion (due to conservation of energy: |work|or|force x distance| is the same on both sides of the pulley)

I think the answer is (c).

So the answer to my question (a) is 98 mm?

You need to be clear about what your variables stand for. Is x the displacement of the mass, or the extension of the spring? Or what? Where is the tension? In S, or in the string connecting the mass to the pulley?

You didn't say what Ts and Ts' are. I'd just stick to one T for tension in any one place, but realize that it is a variable.

x = extension of the spring measured from its unstretched length
Ts = tension in wire S when the mass is in equilibrium position and the spring extends by 98 mm
Ts' = tension in wire S when the mass is released (after being pushed upwards)

Let say T = tension in wire S + restoring force of spring = tension in the string connecting the mass and pulley

You need to consider just one position where the mass is not at equilibrium.
(The other position is just to allow you to calculate the mass.)

When the mass is in equilibrium point, can I say the tension in S is equal to restoring force of spring?

Thanks

Homework Helper
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So the answer to my question (a) is 98 mm?
Yes.
When the mass is in equilibrium point, can I say the tension in S is equal to restoring force of spring?
That is always true, not just at the equilibrium point.

• songoku
Homework Helper
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Yes.
Something looks strange here, you (and the OP) treat the problem like the harmonic motion will be like the mass M was attached to the spring and there was no pulley and no string.
That is always true, not just at the equilibrium point.
Can you expand on this please?

• songoku
Homework Helper
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you (and the OP) treat the problem like the harmonic motion will be like the mass M was attached to the spring and there was no pulley and no string.
No, that would give 196mm.
Read the question carefully. We are told that when the mass is raised to where the spring is relaxed, the spring shrinks by 98mm, but then we are asked how far the mass will descend on release. How far did the mass rise in the first place?
Can you expand on this please?
Since the pulley and string are of negligible mass, the tension in the string always equals the tension in the spring.

Last edited:
• songoku and Merlin3189
songoku
I still don't get equation of simple harmonic motion.

Let say the mass has been released and it moves downwards until the spring extended by 98 mm, the free body diagram of the mass will be: tension T (between mass and pulley) acting upwards and weight acting downwards.

Equation of motion:
W - T = M.a
M.g - T = M.a
M.g - 2.k.x = M.a → from this equation, acceleration is not directly proportional to the displacement

Is the motion of M considered simple harmonic motion?

Thanks

Homework Helper
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I am not sure at all here but my opinion:

M does indeed simple harmonic motion but its amplitude is half of the maximum compression (or maximum extension) of spring from its relaxed position and for this is responsible the pulley.

• songoku and Merlin3189
Homework Helper
Gold Member
Yeah, Delta's nailed the point. You're measuring displacement from the unextended position of the spring. That's why you've got an unwanted constant term
The restoring force needs to be proportional to the displacement from the equilibrium position.

You say
M.g - 2.k.x = M.a
and you know there is no accn. at equilibrium point. If your x were measuring from there, you'd have
M.g - 2.k.0 = M.0 or M.g = 0 obviously not right.

• songoku and Delta2
songoku
Yeah, Delta's nailed the point. You're measuring displacement from the unextended position of the spring. That's why you've got an unwanted constant term
The restoring force needs to be proportional to the displacement from the equilibrium position.

You say
M.g - 2.k.x = M.a
and you know there is no accn. at equilibrium point. If your x were measuring from there, you'd have
M.g - 2.k.0 = M.0 or M.g = 0 obviously not right.

Yeah, I kept messing up between spring extension and displacement of simple harmonic motion.

Now I got the correct equation, along with answer for question (b), which is the same as answer key.

Thank you very much for the help Merlin3189, haruspex and Delta2

• Delta2
Homework Helper
Gold Member
One thing that might be helpful, is to realize that the spring doesn't have to be the spring, just something that behaves like a spring. So here the light frictionless pulley, wire and spring behaves just like a spring, with a different constant.

You can work out how much the force changes as the mass moves. That gives you the spring constant of your equivalent spring.
Here when the mass moves 1 mm, the spring stetches 2 mm, requiring 1 N tension in the wire, so via the pulley, 2 N at the mass. So the imaginary spring is 2 N per 1 mm, giving K = 2000 N/m

• songoku and Delta2
songoku
One thing that might be helpful, is to realize that the spring doesn't have to be the spring, just something that behaves like a spring.
View attachment 263065
So here the light frictionless pulley, wire and spring behaves just like a spring, with a different constant.

You can work out how much the force changes as the mass moves. That gives you the spring constant of your equivalent spring.
Here when the mass moves 1 mm, the spring stetches 2 mm, requiring 1 N tension in the wire, so via the pulley, 2 N at the mass. So the imaginary spring is 2 N per 1 mm, giving K = 2000 N/m

By seeing your working, I just realized there is mistake in my working. I got effective spring constant = 1000 N/m when it should be 2000 N/m. The mistake lies in my equation relating extension of spring, amplitude and displacement from equilibrium position.

Thank you very much Merlin3189