# In a vertical spring, why is the amplitude $2mg/k$ and not $\sqrt{2mg/k} - mg/k$?

When the spring is in unstretched position, the extension is $$0$$. At the equilibrium, which is the mean position of the SHM, extension is $$\frac{mg}{k}$$.

The maximum extension possible should be, by conservation of Energy, $$mgx=\frac{1}{2}kx^2\quad x = \text{extension of spring}$$ and $$x$$ should come out to be $$\sqrt{\frac{2mg}{k}}$$.

Since the origin is at unstretched position and not the mean position,the amplitude should come out to be $$A = \sqrt{\frac{2mg}{k}} - \frac{mg}{k},$$ the displacement between mean position and extreme position.

The books I have seen have $$\frac{mg}{k}$$ as the amplitude. Doesn't it mean that the total extension of spring is $$\frac{2mg}{k}$$ (mean position + extreme position)? Won't that violate the law of conservation of mechanical energy, as it is greater than $$\sqrt{\frac{2mg}{k}}$$? How is this the amplitude, and what's the logic behind it?

• If $mg/k$ has the dimensions of a length, $\sqrt{2mg/k}$ cannot also have the dimensions of a length. – G. Smith Nov 20 '18 at 6:14
• @G.Smith both are dimensionally consistent. – Yuvraj Singh Nov 20 '18 at 6:49
• Got it man. Thanks . I was making a bad,bad,bad mistake. – Yuvraj Singh Nov 20 '18 at 6:59
• I’m glad you found your mistake, but before you did you thought those two terms were dimensionally consistent. Do you understand that $X$ and $\sqrt{X}$ can be dimensionally consistent only if they are dimension-less, and then they can’t be a length? – G. Smith Nov 20 '18 at 17:35
• @G.Smith Thanks, firstly. Cutting to the point,yes,I realise the facts you stated. High-school student like me sometimes go overboard and think dimensional analysis is for rookies(since it easy),and sometimes make realy ridiculous mistake while neglecting dimensions. Thanks again – Yuvraj Singh Nov 21 '18 at 14:52

## 1 Answer

You said:

"and x should come out to be $$(\sqrt{\frac{2mg}{k}})$$".

It is wrong, by simple algebra; (divide both sides by x) it would be: $$({\frac{2mg}{k}})$$. Now you can see that it do not violate the law of conservation of mechanical energy!