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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?

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    $\begingroup$ If $mg/k$ has the dimensions of a length, $\sqrt{2mg/k}$ cannot also have the dimensions of a length. $\endgroup$ – G. Smith Nov 20 '18 at 6:14
  • $\begingroup$ @G.Smith both are dimensionally consistent. $\endgroup$ – Yuvraj Singh Nov 20 '18 at 6:49
  • $\begingroup$ Got it man. Thanks . I was making a bad,bad,bad mistake. $\endgroup$ – Yuvraj Singh Nov 20 '18 at 6:59
  • $\begingroup$ 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? $\endgroup$ – G. Smith Nov 20 '18 at 17:35
  • $\begingroup$ @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 $\endgroup$ – Yuvraj Singh Nov 21 '18 at 14:52
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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!

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