proof with differential equation

Question

How to prove the last equation from the first one. If I integrate both side there is something left side in terms of arc sine and on left side there is t. then how to prove the equation that is given in terms of cos???

Answer 1

Given equation is from standard spring-mass harmonic motion. Energy conservation KE+ PE

$$ \dfrac12 m \dot x^2 + k \dfrac{x^2}{2} =E \tag 1$$

which is the same as the given equation

$$\dot x^2 = \dfrac{2E}{m}- \dfrac{k x^2}{m} \tag 2 $$

Differentiate wrt time $t$ gives force equilibrium

$$ m \ddot x + k x= 0 \tag3 $$

having solution $ \omega^2 = k/m $

$$ x = x(t) = A \sin \omega t + B \cos \omega t \tag 4 $$

To find phase let

$$ \dfrac{A}{\sqrt{A^2+B^2}} =\sin K ,\dfrac{B}{\sqrt{A^2+B^2}} = \cos K, \tan^{-1}\dfrac{A}{B} =K, x_{max}= {\sqrt{A^2+B^2}} \tag 5 $$

to write

$$ x = x_{max} \cos(\omega t - K) \tag 6 $$

From(1) at extremity maximum potential energy at $ t=0, x_{max}=\sqrt{ \dfrac{2E}{k}} \tag 7$

so that when phase difference $K$ is included

$$ x =\sqrt{ \dfrac{2E}{k}} \cos(\omega t + K). \tag 8 $$

The sign of arbitrary constant $K$ is also arbitrary. There is a printo in the given text at 4. , missing radical sign of $ \sqrt{k/m} $.

Answer 2

I may have done something dumb, but the below gives almost the answer given with the exception that there is an extra square root in the solution. I'll leave it to someone to find my error.

The left-hand side can be rewritten to be the derivative of the $\arccos()$ function. A standard differential is: $$d\left(\arccos\left(\frac{u}{n}\right)\right) = -\frac{du}{\sqrt{n^2 - u^2}}$$ If $E$, $m$, and $k$ are constants, then we can simply say: $$n^2 = \frac{2E}{m}\quad\quad u = x\sqrt{\frac{k}{m}} \quad\quad du = \sqrt{\frac{k}{m}}\,dx\quad\quad dx = \frac{du}{\sqrt{\frac{k}{m}}}$$

That means: $$ \int -\frac{\frac{du}{\sqrt{\frac{k}{m}}}}{\sqrt{n^2 - u^2}} = \pm\int dt \\ \int -\frac{du}{\sqrt{n^2 - u^2}} = \pm \sqrt{\frac{k}{m}} \int dt \\ \arccos\left(\frac{u}{n}\right) = \sqrt{\frac{k}{m}} t + K \\ \cos\left(\arccos\left(\frac{u}{n}\right)\right) = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\ \frac{u}{n} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) $$ A few notes: (1) in the first step I added a negative sign. Since the right was $\pm$ anyway, I could multiply both sides by $-1$. Then, I got rid of the $\pm$ once I was multiplying by a square root, since those are $\pm$ by definition anyway.

Now we can undo our substitutions and get: $$ x \frac{\sqrt{\frac{k}{m}}}{\sqrt{\frac{2E}{m}}} = \cos\left( \sqrt{\frac{k}{m}} t + K\right) \\ x \sqrt{\frac{\frac{k}{m}}{\frac{2E}{m}}} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\ x \sqrt{\frac{k}{m}\frac{m}{2E}} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\ x \sqrt{\frac{k}{2E}} = \cos\left(\sqrt{\frac{k}{m}} t + K\right) \\ x = \sqrt{\frac{2E}{k}} \cos\left(\sqrt{\frac{k}{m}} t + K\right)$$

This is the solution in the original question, but with an extra square root. Not sure if I went wrong somewhere or the book or the questioner.