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Consider the following problem

I pull a mass m resting at x = 0 on a frictionless table connected to a spring with some k by an amount A and let it go. What will be its speed at x=0?

I know how to solve it using the law of energy conservation but as a challenge I wanted to find the solution without it.

So from the 2nd Newton's law it's obvious that a = k*x/m. But I don't know how to integrate it in order to find velocity since a = f(x(t)).

How would I solve this without using the law of energy conservation?

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Your system is a harmonic oscillator. Have a look at and if you run into problems with the maths ask again here. – John Rennie Oct 26 '12 at 8:08
up vote 0 down vote accepted

I believe you could use some differential calculus.
A helpful formula is a = d(v^2 / 2)/dx, derived via a = dv/dt = dv/dx * dx/dt

$$a = {-kx \over m}$$ $$a = {\operatorname{d}\!({1 \over 2}v^2)\over\operatorname{d}\!x}$$ $$\therefore -{1 \over 2}v^2 = \int {k \over m} x dx = {k \over 2m}x^2+c$$ When $x=A$, $v=0$: $\therefore c = +{kA^2 \over 2m}$ $$\therefore v = +\sqrt{{-kx^2+kA^2 \over m}}$$ When $x=0$, $v=\sqrt{{kA^2 \over m}}$

This is the same result as that you would get using energy.

(I've assumed that the force causing the spring to retract is F = -kx.
I've made this assumption based on it being the force required to stretch the string at constant velocity)

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If you know $a(x)$ then use the following formulas ( derived from $a = \frac{{\rm d}v}{{\rm d}x} v$ )

$$ \frac{1}{2} v(x)^2 = \int a(x)\,{\rm d}x + C_1 $$

$$ t = \int \frac{1}{v(x)}\,{\rm d}x + C_2 $$

where the constants $C_1$ and $C_2$ are found from the initial conditions.


$a(x)=-\frac{k}{m} x$

$$ \frac{1}{2} v(x)^2 = \int \text{-}\frac{k}{m} x \,{\rm d}x + C_1 = \frac{k\, (A^2-x^2)}{2\,m} $$ for $v=0$ at $x=A$. The above is inverted to

$$ v(x) = \sqrt{ \frac{k\,(A^2-x^2)}{2 m} } $$

The solution is then $v(0) = \sqrt{ \frac{k\,A^2}{2 m} } $

The time is more complex to get

$$ t = \int \frac{1}{v(x)}\,{\rm d}x + C_2 = C_2 + \int \frac{ \sqrt{ \frac{m}{k} } }{ \sqrt{A^2-x^2} } \,{\rm d}x = \sqrt{ \frac{m}{k}} \left( \arccos\left(\frac{x}{A}\right) \right) $$

for $x=A$ at $t=0$. This is inverted for

$$ x(t) = A \cos\left(\sqrt{\frac{k}{m}}\,t\right) $$

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I just found this great link that does exactly this: CodeCogs/Reference – ja72 Oct 26 '12 at 17:49

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