Question

An object of mass M is attached to a vertical spring which slowly lowers to equilibrium position. This extends the spring x. If the same objects is attached to the same vertical spring but permitted to fall suddenly instead, then the maximum extension of the spring is?

• A. x
• B. $2x$
• C. $\frac{x}{2}$
• D. $3x$

Answer 1

The correct option is B $2x$

when object is taken slowly to equilibrium position then $mg=kx$ is valid.

gives $x=\frac{mg}{k}$

but when released form natural length of spring then maximum extension will be calculate by balancing energy, decrease in potential energy will be equal to the energy stored in the spring.

$mg{x}_2=\frac{1}{2}k{x}_2^2$, $=>x_2=\frac{2mg}{k}$4

so $x_2=2x$

so maximum extension in second case is 2x.

so best option is option B.