Tension in the strings after string is cut

In summary, the tension in string 1, the cutting of string 2, and the release of string 3 causes the mass of the spring to oscillate. The oscillation gets zero amplitude and frequency as the mass of the spring becomes negligible.
  • #1
Tanya Sharma
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Homework Statement



Q .Find the tensions in the strings (1), (2) and (3) and the acceleration of the mass ‘m’ just after :
(a) string (1) is cut
(b) string (2) is cut
(c) string (3) is cut

Homework Equations


The Attempt at a Solution



I feel the answer should be

(a) T1 = 0 ; T2 = mg ; T3 = 2 mg ; a = g
(b) T1 = mg ; T2 = 0 ; T3 = 0 ;a = 0
(c) T1 = mg ; T2 = 0 ; T3 = 0 ;a = 0

I would be grateful if somebody could check my understanding of the problem . Have I answered correctly ?
 

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  • #2
When string 2 is cut, why do you say the mass won't start falling?

What properties are you attributing to the spring?
 
  • #3
NascentOxygen said:
When string 2 is cut, why do you say the mass won't start falling?

What properties are you attributing to the spring?

The extension in the spring will not change instantaneously .Just after the string 2 is cut,the mass will still have upwards force T1 and mg downwards ,with kx=T1=mg ,which we had obtained from the equilibrium state (just before the string is cut) .
 
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  • #4
Tanya Sharma said:

Homework Statement



Q .Find the tensions in the strings (1), (2) and (3) and the acceleration of the mass ‘m’ just after :
(a) string (1) is cut
(b) string (2) is cut
(c) string (3) is cut


Homework Equations





The Attempt at a Solution



I feel the answer should be

(a) T1 = 0 ; T2 = mg ; T3 = 2 mg ; a = g
(b) T1 = mg ; T2 = 0 ; T3 = 0 ;a = 0
(c) T1 = mg ; T2 = 0 ; T3 = 0 ;a = 0
There are some arbitrary assumptions that can be made here, and, as always, it's prudent that you fall in line with the view of whoever will be marking your work. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

It appears that you are attributing to the spring a significant mass, i.e., the spring has mass which is not insignificant under dynamic conditions. Accordingly, I agree with your answers for (a) and (b).

Are we likewise to assume the pulley has mass which is not insignificant under dynamic conditions? If not, why not? If so, then I don't agree with your answer (c).

Your answer for (c) would accord with the pulley being massless, without inertia.

I'm hoping someone else will confirm or refute my answer, as I'm not confident answering questions which seem to begin, "First, flip a coin ...".
 
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  • #5
NascentOxygen said:
There are some arbitrary assumptions that can be made here, and, as always, it's prudent that you fall in line with the view of whoever will be marking your work. http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif

It appears that you are attributing to the spring a significant mass, i.e., the spring has mass which is not insignificant under dynamic conditions. Accordingly, I agree with your answers for (a) and (b).

Are we likewise to assume the pulley has mass which is not insignificant under dynamic conditions? If not, why not? If so, then I don't agree with your answer (c).

Your answer for (c) would accord with the pulley being massless, without inertia.

I'm hoping someone else will confirm or refute my answer, as I'm not confident answering questions which seem to begin, "First, flip a coin ...".

I think the pulley and the spring are assumed massless in this problem.Do you mean a) and b) are correct if spring has mass and c) is correct if pulley is massless ?

Could you please explain how the mass of spring affects our results ?
 
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  • #6
Imagine you fasten the spring to a wall, the other end to a small ball. Pull out the ball and then release on a horizontal surface, what happens? The ball will oscillate.
The same happens if you fasten balls at both ends. Pull the balls apart, then release them. They will oscillate about the common centre of mass.

Choose one mass smaller and smaller. The centre of mass gets closer and closer to the centre of the other ball, which will oscillate with less and less amplitude. At the limit, the amplitude of the oscillation gets zero, while the frequency stays finite. The acceleration of the ball becomes zero.

Here the spring was considered massless, but you get similar result if you replace the other mass with a massive spring. There is initially tension along the spring. That tension acts on the mass at the end. Releasing the other end, the spring-ball system start to do SHM. The spring contracts, but the ball feels the initial tension so accelerates inward, toward the CM of the system.

If you release the free end of the spring, the tension at the free end becomes zero and a wave is initiated. The speed of the wave depends on the mass of unit length of the spring. If that mass is zero, the speed is infinite. The disturbance reaches the other end in no time.

ehild
 
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  • #7
Thanks ehild :smile:

ehild said:
If you release the free end of the spring, the tension at the free end becomes zero and a wave is initiated. The speed of the wave depends on the mass of unit length of the spring. If that mass is zero, the speed is infinite. The disturbance reaches the other end in no time.

ehild

Is it correct to say that conclusions(answers) drawn in post#1 are correct under the assumption that spring is not massless and pulley is massless ?
 
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  • #8
Tanya Sharma said:
The extension in the spring will not change instantaneously .
Why not? Ask yourself "why can't the spring instantly revert to being unstretched?" The only answer can be "because the coils of the spring have mass, therefore inertia", and it's this that slows down the spring's movement.

Just after the string 2 is cut,the mass will still have upwards force T1 and mg downwards ,with kx=T1=mg ,which we had obtained from the equilibrium state (just before the string is cut) .
The only basis on which the spring is able to exert any force on the mass below it once string 2 is cut must be inertia due to the spring coils themselves having mass. When string 2 is cut, the two ends of the stretched spring both tend to retract towards the middle of the spring, and it's this upward pull by the lower end of the spring which explains how tension persists in string 1 after string 2 is cut. If the spring were massless, tension in string 1 would instantly fall to zero simultaneously with the cutting of string 2.

I'm undecided on for how long that tension in string 1 will equal mg. It would make for an interesting classroom demonstration to verify the initial magnitude of this tension and its variation--if any--with time.
 
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  • #9
Have I concluded correctly in post#7 ?
 
  • #10
Tanya Sharma said:
Is it correct to say that conclusions(answers) drawn in post#1 are correct under the assumption thst spring is not massless and pulley is massless ?
Yes, that's my conclusion.
 
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  • #11
Wonderful...Thank you very much Nascent :smile:

Is it correct to say that the massless spring is equivalent to a massless string as far as transmission of tension is concerned across its length ?
 

FAQ: Tension in the strings after string is cut

How does cutting a string affect the tension in the remaining strings?

When a string is cut, the tension in the remaining strings increases. This is because the weight of the string that was cut is no longer pulling down on the remaining strings, causing them to become tighter.

Does the tension in the strings decrease if the string is cut at an angle?

Yes, the tension in the strings will decrease if the string is cut at an angle. This is because the weight of the cut string is distributed between the remaining strings, causing each one to have less tension.

Why does the tension in the strings increase when a string is cut?

The tension in the strings increases when a string is cut because the weight of the cut string is no longer pulling down on the remaining strings, causing them to become tighter to compensate for the loss of weight.

Is the tension in the strings affected by the thickness or material of the string?

Yes, the tension in the strings can be affected by the thickness and material of the string. Thicker and more rigid strings will have a higher tension than thinner and more flexible strings.

Can the tension in the strings be measured after a string is cut?

Yes, the tension in the strings can be measured after a string is cut using a tension meter or by calculating the amount of weight that is needed to pull the string taut. However, the tension may be slightly different from the original tension before the string was cut due to factors such as elasticity and friction.

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