Question

An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amount d. If the same object is attached to the same vertical spring but permitted to fall instead, through what maximum distance does it stretch the spring? Please show all work. Step by step.

Answer 1

Answer:

2d

Explanation:

For any instance equivalent force acting on the body is

$mg-kd= m\frac{d}{dt}\frac{dx}{dt}$

Where

m is the mass of the object

k is the force constant of the spring

d is the extension in the spring

and

d/dt(dx/dt)=  is the acceleration of the object

solving the above equation we get

$x= Asin\omega t +d$

where

$\omega= \sqrt{\frac{k}{m} } = \frac{2\pi}{T}$

A is the amplitude of oscillation from the mean position.

k= spring constant , T= time period

Here  we are assuming  that at t=T/4

x= 0   since, no extension in the spring

then

A=- d

Hence

x=- d sin wt + d

now, x is maximum when sin wt=- 1

Therefore,

x(maximum)=2d