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    With these questions I'm struggling to figure out exactly which value to use as x in the T = \frac{\lambda x}{l} formula

    Because the string is already stretched, how does this factor into the value for x? Does the extension of moving the particle P \frac{1}{2}L from where it is at rest (it is attached to the centre) compound with the initial extension to give x_1 = x + L?

    Many thanks

    Question is from Edexcel M4 January 2004 Q4

    Tagged some people who may be able to help: ghostwalker tiny hobbit Gome44
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    (Original post by Euclidean)
    Attachment 517837

    With these questions I'm struggling to figure out exactly which value to use as x in the T = \frac{\lambda x}{l} formula

    Because the string is already stretched, how does this factor into the value for x? Does the extension of moving the particle P \frac{1}{2}L from where it is at rest (it is attached to the centre) compound with the initial extension to give x_1 = x + L?

    Many thanks

    Question is from Edexcel M4 January 2004 Q4

    Tagged some people who may be able to help: ghostwalker tiny hobbit Gome44
    link doesnt work
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    (Original post by Gome44)
    link doesnt work
    This one should work:

    Name:  Screen Shot 2016-04-01 at 16.17.09.png
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    (Original post by Euclidean)
    This one should work:

    Name:  Screen Shot 2016-04-01 at 16.17.09.png
Views: 49
Size:  159.7 KB

    Draw a diagram. The length of the string from P to B is 1.5L-x. The natural length of this string is L, thus extension is 0.5L-x

    So yes you do factor in the initial extension
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    (Original post by Gome44)
    Draw a diagram. The length of the string from P to B is 1.5L-x. The natural length of this string is L, thus extension is 0.5L-x

    So yes you do factor in the initial extension
    Thanks for explaining, I've got that tension right now.

    Just checked the solution and I can't see why they have included a second tension T_2 on their answer though?

    Name:  Screen Shot 2016-04-01 at 16.59.36.png
Views: 43
Size:  95.2 KB

    Surely if you pull this particle on a taut string to the right there will only be a tension to the left opposing that extension, why is there a tension T_2 in this diagram?

    Edit: Is this about the initial tension from the stretch from A to B. We assume that the string is in two parts, one with extension \frac{1}{2} L and another with the same extension? Hence when we factor in the tension to one string we have to do the same for the other?
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    (Original post by Euclidean)
    Thanks for explaining, I've got that tension right now.

    Just checked the solution and I can't see why they have included a second tension T_2 on their answer though?

    Name:  Screen Shot 2016-04-01 at 16.59.36.png
Views: 43
Size:  95.2 KB

    Surely if you pull this particle on a taut string to the right there will only be a tension to the left opposing that extension, why is there a tension T_2 in this diagram?

    Edit: Is this about the initial tension from the stretch from A to B. We assume that the string is in two parts, one with extension \frac{1}{2} L and another with the same extension? Hence when we factor in the tension to one string we have to do the same for the other?
    Once the particle is moving, both halves of the string will be longer than their natural length, so both will have a tension.
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    (Original post by tiny hobbit)
    Once the particle is moving, both halves of the string will be longer than their natural length, so both will have a tension.
    Thanks for the explanation, it makes sense now :doh:
 
 
 
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