# Extension of a string One end of a light elastic string of natural length `l` , passing through a small smooth ring of mass `m` , is attached to a point `O` on the ceiling of a room. A particle...

**Extension of a string**

One end of a light elastic string of natural length `l` , passing through a small smooth ring of mass `m` , is attached to a point `O` on the ceiling of a room. A particle `P` of mass` M` attached to the other end of the string hangs in equilibrium, with the ring being held at rest at the point `O` . If `2Mg` is the modulus of elasticity of the string,

Q: show that the extension of the string in the equilibrium position is `l /2` .

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### 1 Answer

We have a particle P of mass M suspended on a string of natural length l with a modulus of elasticity 2Mg.

When the 'system' (the set up of the string, the particle and the ring) is in equilibrium, ie the particle P is at rest, the resultant force acting on it is zero (according to *Newton's Second Law*). Therefore, the downward weight of the particle should balance the upward force exerted by the string on the particle (which is equal in *magnitude* to the tension T of the string).

Now, *Hooke's Law* states that the extension x of the string is proportional to the tension T applied to the string. More specifically

`T = (lambda/l)x`

where `lambda` is the modulus of elasticity of the string (2Mg here) and l is the natural length. Since T must be equal to the downward force on the string (the force is equal and in the opposite direction according to *Newton's Third Law*, which is Mg (mass of P x acceleration due to gravity - *Newton's Second Law F = ma*) then *Hooke's Law* gives that

`Mg = ((2Mg)/l)x` which implies that

`x = l/2`

**Answer: using Hooke's Law and Newton's Second and Third Laws, x = l/2 is the extension of the string when the 'system' is in equilibrium.**