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

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. Upon inspecting this problem it doesn't explicitly mentions about the height from where the bob is hanging.

But I'm assuming that the intended elongation for the spring is $$y$$ hence it appears in the alternatives. Therefore, can someone help me here?.

• This question would be more appropriate for the Physics Stack Exchange. Mar 25 '20 at 9:54

The first thing to note is that in all of options, when $$y=0$$ then $$v=0$$. So the extension of the spring is measured as a displacement from a point where the oscillating mass is stationary.

The second thing to note is that an expression for the energy of the system must contain three terms:

1. Kinetic energy which is $$\frac 1 2 mv^2$$.
2. Gravitational potential energy which is $$-mgy$$ (since $$y$$ is described as the elongation of the spring we can assume that $$y$$ increases in the downwards direction).
3. Potential energy stored in the spring, which is $$\frac 1 2 Ky^2$$ where $$K$$ is the spring constant and $$y$$ is measured as a displacement from the normal length of the spring (without the mass).

Adding these three terms together then conservation of energy gives us:

$$\frac 1 2 mv^2 -mgy + \frac 1 2 Ky^2 = \text{constant}$$

If we assume that the mass has been released when the spring is at its natural length, then $$v=0$$ when $$y=0$$, so

$$\frac 1 2 mv^2 -mgy + \frac 1 2 Ky^2 = 0\\ \Rightarrow \frac 1 2 mv^2 = mgy - \frac 1 2 Ky^2 \\ \Rightarrow v^2 = 2gy -\frac K m y^2$$