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:


Therefore in this situation it would be:


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?.

  • $\begingroup$ This question would be more appropriate for the Physics Stack Exchange. $\endgroup$
    – Cesareo
    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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.