# 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` .

mathsworkmusic | (Level 2) Educator

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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.