Question

An object attached to a vertical spring is slowly lowered to its equilibrium position, thus stretching the spring by an amount $d$. Now the same object attached to the vertical spring is allowed to fall. This will stretch the spring by

• A. $d/2$
• B. $d/4$
• C. $d$
• D. $2d$

Answer 1

The correct option is C $d$

When the mass is slowly lowered the spring will extend till the gravitational force on the mass is balanced by the restoring force of the spring. Such that,

$mg=kd$

$\frac{mg}{k}=d$

When the mass is allowed to fall freely it will attain a kinetic energy which will put the mass in an oscillatory about the mean extension of the spring which is also $d$.