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$\newcommand{\d}{\mathrm{d}}$I am having a problem here because acceleration varies with distance, and I am not able to relate it with time

\begin{align} a &=-\frac{kx}{m} \\ v\frac{\d v}{\d x} &= -\frac{kx}{m} \end{align} For $x_{\max}$ \begin{align} 0-v_{0}^2 &= -\frac{kx^2}{m} \\ x &= \sqrt{\frac{mv_0}{k}} \end{align}

But I still can’t find time

Nathanael Skrepek
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Aditya
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    You have been given an example of simple harmonic motion https://en.wikipedia.org/wiki/Simple_harmonic_motion#:~:text=In%20mechanics%20and%20physics%2C%20simple,of%2C%20the%20object's%20displacement%20vector. – Simon Terrington Jul 21 '20 at 10:31
  • Hint: solve line 2 to get v in terms of x. Then substitute v = dx/dt and solve for x in terms of t. – Julia Hayward Jul 21 '20 at 10:46
  • Well, this is a math discussion area..... but never mind - questions are inevitable..... – Spectre Jul 21 '20 at 10:48

1 Answers1

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$\newcommand{\d}{\mathrm{d}}$We have, $$ v \,\d v =-Cx \,\d x \\ \int_{v_0}^{v(x)} v \,\d v = -C\int_0^x x\,\d x \\ \frac{v(x)^2-v_0^2}{2}=-Cx \\ v(x)=\sqrt{v_0^2-2Cx}, $$ where $C=\frac km$. Using $v=\frac{\d x}{\d t}$, $$\frac{\d x}{\sqrt{v_0^2-2Cx}} = \d t$$ Now just integrate both sides to get an expression for $x(t)$, after which you need to solve $x(t)=0$.

Nathanael Skrepek
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Vishu
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