[meatball893]

I was doing some revision for M3 (Edexcel) and this question (under Further Dynamics: Vertical oscillations of a particle attached to an elastic string or spring) has me stumped.

Heinemann Book M3, P92-3, Q11

A light string has two ends fastened together to form a loop of natural length and modulus . The string is placed over two smooth fixed pegs A and B where AB is horizontal and of length . A particle P of mass is attached to the mid-point of the lower part of the loop and hangs in equilibrium vertically below C, the mid-point of AB. Explain why the extension of the string is equal to twice the length of AP.

I have tried for quite a while on this, but can't make any headway. I'm sure it's quite simple, which makes it rather embarrassing...

[benwellsday]

I'm not sure if this is what it wants but I think:-
Since one end of the loop is just horizontal along AB, and hence 2L long, the rest of the loop must be the extension.
Hence the bit with the particle hanging is all extension.
Then AP is the same length as BP, since if it wasn't there wouldn't be equilibrium in the horizontal directions (try considering an arbitrary case and showing that the horizontal tensions only cancel when the length of the string along AP is equal to that of BP, that should work).
I'll look at it now and see if I can show that, but it makes sense intuitively right?

[ghostwalker]

Originally Posted by meatball893

I was doing some revision for M3 (Edexcel) and this question (under Further Dynamics: Vertical oscillations of a particle attached to an elastic string or spring) has me stumped.

Heinemann Book M3, P92-3, Q11

A light string has two ends fastened together to form a loop of natural length and modulus . The string is placed over two smooth fixed pegs A and B where AB is horizontal and of length . A particle P of mass is attached to the mid-point of the lower part of the loop and hangs in equilibrium vertically below C, the mid-point of AB. Explain why the extension of the string is equal to twice the length of AP.

I have tried for quite a while on this, but can't make any headway. I'm sure it's quite simple, which makes it rather embarrassing...

Yep, it's quite simple:

The natural length of the loop is 2l (this may be what's causing you a problem. The length is measured from starting at one point on the loop and going round until you come back to the start. NOT by holding it laid out in a line and the distance between the two ends)

When stretched across the two pegs with the weight below the total length will be (look at your diagram) AB + BP + AP. By symmetry AP=BP, and you are told AB = 2l

So. stretched length is 2l + 2AP. Hence extension = 2l +2AP - 2l (the original length) = 2AP.

[meatball893]

Ah, I see. I was thinking about it as having length 2l when laid in a line, so the original string had length 4l. Silly me.

Thanks guys!