# How to find the velocity as a function of the elongation of a spring when it is hanging from a ceiling?

#### Chemist116

The problem is as follows:

A mass whose mass is $m$ is hanging vertically from a ceiling which is tied to a spring which has a constant of $K$ is oscillating. Given this condition find the velocity as a function of the elongation of the spring.

The alternatives given are as follows:

$\begin{array}{ll} 1.&\sqrt{\frac{K}{m}y^2+2gy}\\ 2.&\sqrt{2gy-\frac{K}{m}y^2}\\ 3.&\sqrt{\frac{K}{m}y^2-2gy}\\ 4.&\sqrt{\frac{K}{m}y}\\ 5.&\sqrt{2gy}\\ \end{array}$

How exactly should I find the velocity in this situation?. Could it be that since appears a square root that is related to the conservation of mechanical energy?

If this is the case it would be that:

$\frac{1}{2}ky^2=\frac{1}{2}mv^2$

Therefore in this situation it would be:

$v=\sqrt{\frac{ky^2}{m}}$

But it doesn't appear in any of the alternatives. Exactly which part did I missunderstood. Can someone help me here?. Upon inspecting this problem it doesn't explicitly mentions anything regarding the height, but I'm assuming that the elongation is $y$ and it might be the intended meaning in this problem. But as indicated it doesn't check with any of the alternatives. Can anyone help me here please?.

Last edited:

#### skeeter

Math Team
Note that as the mass falls stretching the spring, three forms of mechanical energy are changing.
At the very top, gravitational potential energy is a maximum, i.e. $U_g = E_T$, while elastic potential energy, $U_e$, and kinetic energy, $KE$, are both zero.
At the very bottom where the spring is stretched to its maximum, $U_g = 0$, $KE=0$, and $U_e = E_T$

Assuming $y$ represents the magnitude of spring stretch and the mass is released from a position where the spring is at its natural length ...

$U_g = E_T - mgy$
$U_e = \dfrac{1}{2}ky^2$
$KE = \dfrac{1}{2}mv^2$

The sum of all three forms of mechanical energy, $U_g, \, U_e, \, \text{ and } KE$ is total energy, $E_T$ ...

$E_T = E_T - mgy + \dfrac{1}{2}ky^2 + \dfrac{1}{2}mv^2$

$mgy - \dfrac{1}{2}ky^2 = \dfrac{1}{2}mv^2$

$\sqrt{2gy - \dfrac{k}{m}y^2} = v$

• Chemist116, topsquark and idontknow

#### Chemist116

Note that as the mass falls stretching the spring, three forms of mechanical energy are changing.
At the very top, gravitational potential energy is a maximum, i.e. $U_g = E_T$, while elastic potential energy, $U_e$, and kinetic energy, $KE$, are both zero.
At the very bottom where the spring is stretched to its maximum, $U_g = 0$, $KE=0$, and $U_e = E_T$

Assuming $y$ represents the magnitude of spring stretch and the mass is released from a position where the spring is at its natural length ...

$U_g = E_T - mgy$
$U_e = \dfrac{1}{2}ky^2$
$KE = \dfrac{1}{2}mv^2$

The sum of all three forms of mechanical energy, $U_g, \, U_e, \, \text{ and } KE$ is total energy, $E_T$ ...

$E_T = E_T - mgy + \dfrac{1}{2}ky^2 + \dfrac{1}{2}mv^2$

$mgy - \dfrac{1}{2}ky^2 = \dfrac{1}{2}mv^2$

$\sqrt{2gy - \dfrac{k}{m}y^2} = v$
Gee I totally overlooked that there were three energies in this setup. In your analysis you're calculating the energy at some point during the elongation of the spring, but not when all the spring has stretched I mean in the bottom. Why? Btw at first I was confused why did you used

$U_g = E_T - mgy$

But then I thought that this was to get the potential energy at some point of the trajectory of the elongation. I just wondered why?. Typically energy analysis is made during initial and final points. Let's say in a rollercoaster.

#### skeeter

Math Team
In the problem solution choices, $y$ represents the amount of spring stretch which increases as the mass moves downward, so $y=0$ is at the top ... which makes $mgy$ zero at the top. Total energy of the system is $U_g$ at the top, and decreases as $y$ increases, so the equation $U_g=E_T-mgy$ correctly represents the gravitational potential energy as a function of stretch distance.

Consider a graph of Energy as a function of stretch for interpretation ... • topsquark