Motion of a point is equal to dynamic harmonic oscilation

[prehisto]

Homework Statement

Point with mass is moving along the positive direction of x axis, its velocity is described by (A-Bx^2)^(1/2). Show that its equation of motion describes dynamic harmonic oscillation and find period (T) of this oscillation.

Homework Equations

v=(A-Bx^2)^(1/2)
A and B is known constants

The Attempt at a Solution

[/B]
Hi, guys!
My first instinct is to try to get something from F=m*a=-k*x.
Further more m*(dv/dt)=-k*x, but immediately i see that its wrong because v=v(x) not v(t).

So ,please, could someone give me some advice.
Thank you!

 

[paralleltransport]

Hint short:

If energy is conserved for a SHO along the lines of
$$ stuff1 \times v^2 + stuff2 \times x^2 = constant$$
no?

Hint, long:

$$ F = m {dv \over dt} = m {dv \over dx} {dx \over dt} = m v {dv \over dx} = -kx$$

Integrate dx both sides, what do you get?

 

[prehisto]

hello!
Thank you for your hints, I had forgotten connections between derivatives.

If I take the equation mv (dv/dx)=-kx
I obtain m*v*dv=-k*x*dx
By integrating m *(v^2)/2+C₁=-k*(x^2)/2+C₂
I suppose I could now substitute v with given equation but if I do so, the equation becomes very messy and I do not think that it could be reduced to something useful.

What do you suggest?

 

[paralleltransport]

Does your equation with the squares look like a conservation equation to you? :D (protip, you could haved used conservation of energy, but instead you derived it like a beast!)

Anyways, if you solve for v from x from that equation, what do you get? What is the equivalent of
$$\omega = \sqrt{k \over m}$$ when you try to match your equation to the given equation $$v = (A-Bx^2)^{1 \over 2}$$

 

[haruspex]

Show that its equation of motion describes dynamic harmonic oscillation and find period (T) of this oscillation.

There is a simpler alternative to showing it satisfies the usual SHM ODE: show it satisfies the solution to an SHM ODE.

 

[prehisto]

Does your equation with the squares look like a conservation equation to you? :D (protip, you could haved used conservation of energy, but instead you derived it like a beast!)

Anyways, if you solve for v from x from that equation, what do you get? What is the equivalent of
$$\omega = \sqrt{k \over m}$$ when you try to match your equation to the given equation $$v = (A-Bx^2)^{1 \over 2}$$

If I solve for v, I obtain [itex]v = ({\frac{\ 2(C1-C2)}{m}} - {\frac{\ k x^2}{m}})^{(1/2)} [/itex]
which looks similar to given equation and if I assume that
[itex]A≡{\frac{\ 2(C1-C2)}{m}} [/itex]
it can be rewritten
[itex]v = (A - ω^2x^2)^{(1/2)} [/itex]
which means that
[itex] B≡ ω^2 [/itex]