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The differential equation $$\ddot x = -4x$$ describes the motion of a particle moving with simple harmonic motion. Find its period, amplitude and greatest velocity if the initial conditions are $x (0) = 0$ and $\dot x (0) = 4$.

I would really appreciate if someone could help out.

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    $\begingroup$ Given $x'' = -4x$, can you solve for $x(t)$? Then all your questions would have immediate answers $\endgroup$ – An aedonist Sep 11 '17 at 10:35
  • $\begingroup$ would $x'=-4xt+t+4$? or $x'=-2x^2+x+4$? $\endgroup$ – Jr. Mathematician Sep 11 '17 at 10:38
  • $\begingroup$ @Jr.Mathematician Nope. Don't forget that $x$ is a function and $t$ is the variable. So you need a function whose second derivative is negative four times what it is. To make it easier do you know any function such that if you differentiate it twice you get negative the function? Think $f^{\prime\prime}(t)=-f(t)$ $\endgroup$ – DRF Sep 11 '17 at 10:48
  • $\begingroup$ $$x(t)= 2 \sin (2 t)$$ $\endgroup$ – Raffaele Sep 11 '17 at 11:47
  • $\begingroup$ It seems surprising to me that someone should get to a point in their studies where they receive this problem to do without ever having seen any formulas for simple harmonic motion (that is, without having seen the solutions to similar problems). In that case, simply looking up the words "simple harmonic motion" would have been a good start. $\endgroup$ – David K Sep 11 '17 at 12:54
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This is a standard Second Order ODE.

$x''=f(x)$ where $f(x)=-4x$

$p=x'$ therefore, $$x''=\frac{dp}{dt}=\frac{dp}{dx} \frac{dx}{dt}=p\frac{dp}{dx}$$

We write: $x''=pdp/dx$

$$p dp =x''dx$$

$$p dp=-4x dx$$

This can be solved in terms of separation of variables.

$$\int pdp=\int-4xdx$$

$$p^2=2(-2x^2)+2c$$

We know that when $x=0, x'=p=4$,

$$c=8$$

$$x'=\sqrt{-4x^2+16}$$

You thus have an equation for the particle's velocity in terms of its position, and can work with this to find the desired values.

$$\int \frac{dx}{\sqrt{16-4x^2}} = \int dt$$

$$t=\frac{1}{2}arcsin(\frac{x}{2})+c_1$$

$c_1=0$ because when $t=0$, $x=0$

$$sin(2t)=\frac{x}{2}$$

$$x=2sin(2t)$$

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  • $\begingroup$ It's wrong. Solution is $x(t)= 2 \sin (2 t)$ $\endgroup$ – Raffaele Sep 11 '17 at 11:48
  • $\begingroup$ It wasn't wrong, I have completed the answer, as it was incomplete before. Apologies, have separated the variables so that position is given as a function of time. $\endgroup$ – Harry Alli Sep 11 '17 at 12:18
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By inspection, the functions with the properties that their second derivatives are a negative multiple of themselves are the sine and cosine. Hence, try $$ x(t)=A\cos\omega t + B\sin\omega\, .\tag{1} $$ They satisfy $\ddot{x}(t)=-4x(t)$ if $\omega^2=4$, i.e. $\omega=+2$ (the positive root is a matter of convention).

Using (1) and $x(0)=0=A$ you find $A=0$ and from $\dot{x}(0)=4=2B$ you get $B=2$. Thus, the solution to your particular problem is $$ x(t)= 2\sin(2t)\, . $$

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