Masoud Masoud

Posted on Monday, 12 May, 2003 - 10:14 pm:

A light string has its two ends fastened together to form a loop of natural length 2l and modulus of elasticity mg. The string is placed over two smooth fixed pegs A and B where AB is horizontaland of length 2l. A particle P of mass 2m 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.

a) Explain why the extension of the string is equal to twice the length of AP.
b) Show that the distance CP is l.
P is projected vertically downwards from its equilibrium position.
c) Show that, providing the string does not break and P does not reach the level of AB, the subsequent motion is simple harmonic.

I would sincerely appreciate some help on parts b and c.
My solution:
a) The string is extended when it is wound around the peg, consequently any movement of the section of string would cause a double extension, that of AP and BP. Those extensions should be the same hence the extension of the whole string is equal to twice the length AP. Is this correct or am I missing something?
Thanks

Arun Iyer

Posted on Tuesday, 13 May, 2003 - 06:53 pm:

for a) consider forces along the strings and the weight of the particle. Resolving the string forces along weight of the particle, you would get some equation like

2 mg \cos θ = mg

$\cos θ = 1 / 2$ and the result simply follows.

for b)
this follows from a.

for c)
the objective here is to show that the resultant force acting on the particle(or the acceleration of the particle) is directly proportional to the displacement.

Think about the above statement and see if you get anywhere to the solution.

love arun

Masoud Masoud

Posted on Tuesday, 13 May, 2003 - 09:14 pm:

Good day,
a) 2mg cos(t) = mg, can you please elaborate on that?
T = mg(x)/l
How can you determine the value of x, according to your expression its "l".

b) How would you prove that CP = l (lenght)

c) how?
Thank you very much for your help, it is much appreciated.
Masoud

Matthew Smith

Posted on Tuesday, 13 May, 2003 - 10:40 pm:

It's worth noting that the extension of the string is always twice AP, regardless of the weight of the mass. This follows from the definition of the extension (extended length minus natural length). So you can do part (a) without thinking about forces at all.

For part (b), you can resolve forces on the particle in terms of some angle $θ$ . Then use the relation of part (a) to find an equation for the tension in the string. You also need a trigonometric relation between x=AP and l. These three equations can be solved to work out $θ$ , and once you've got that CP is easy.

$\cos θ$ doesn't seem to be 1/2 in this case: I think Arun only meant it as an example of the sort of equation you will end up with - not the actual answer to the problem! Matthew.

Masoud Masoud

Posted on Wednesday, 14 May, 2003 - 07:49 am:

Good morning everyone,
Thanks a lot for your help, I really appreciate it.
Masoud