Question

A spring is hung vertically from a fixed support. When an object of mass m is attached to the end of the spring, it stretches by a distance y. When an object of mass of 2m is hung from the spring, it stretches by a distance 2y. A second, identical spring is then attached to the free end of the first spring.

If the object of mass 2m is attached to the bottom of the second spring, how far will the bottom of the second spring move downward from its outstretched position? Assume the masses of the springs are negligible when compared to m.

Answer 1

Spring force

F = k*x

where F is the force and x is the dispalcement

here displacement is y and force is mg (weight of the object) in downward direction

mg = k y

k = mg/y

when we use 2m mass it displaced by 2y

2mg = k (2y)

k = mg/y

Now two spring are in series

For series conbination we can calculte equivalent spring constant

$\frac{1}{k_{eq}}=\frac{1}{k}+\frac{1}{k}$

$\frac{1}{k_{eq}}=\frac{2}{k}$

${k_{eq}}=\frac{k}{2}$

${k_{eq}}=\frac{mg}{2y}$

2m mass is hanged with this combbination of spring

Force acting in downward direction is 2mg

F = kx

2mg = (mg/2y)(Y)

Y = 4y