# Time to Max Compression of a Spring

1. May 16, 2017

### kq6up

1. The problem statement, all variables and given/known data

A block of mass m slides down an incline with a height h. Later, it collides with a spring and compresses the spring to some maximum displacement. How long does it take to reach maximal compression.

2. Relevant equations

1) $mgh=\frac{1}{2}mv^2+\frac{1}{2}kx^2$
2) $t=\int _{0}^{ x_{max}}{\frac{1}{v}}{dx }$

3. The attempt at a solution

By conservation of energy:

$mgh=\frac{1}{2}mv^2+\frac{1}{2}kx^2$ Rearranging to solve for $v$:

$v=\sqrt{2gh-\frac{k}{m}x^2}$

Inserting into 2):

$t=\int _{0}^{ x_{max}}{\frac{1}{\sqrt{2gh-\frac{k}{m}x^2}}}{dx }$

This is in the form:

$\int { \frac { 1 }{ \sqrt { a^{ 2 }-u^{ 2 } } } } { du }=\sin ^{ -1 } \frac { u }{ a }$

Using the appropriate substitution we get:

$t=\sqrt { \frac { m }{ k } } \left[ \sin^{-1} \frac{u}{a} \right]_{0}^{\sqrt{\frac{k}{m}}x}$

Where: $a=\sqrt{2gh}$ and $x_{max}=\sqrt{\frac{2mgh}{k}}$

Yields the result:

$t=\sqrt{\frac{m}{k}}\frac{\pi}{4}$

Does this look correct? The result is independent of gravity and height of the initial mass.

Thanks,
kQ6Up

Last edited by a moderator: May 16, 2017
2. May 16, 2017

### kuruman

Almost correct. Have you studied simple harmonic motion? If so, what fraction of the period is the time from equilibrium to maximum displacement?

3. May 16, 2017

### kq6up

Ah, I see that it would be $\frac{T}{4}$, and that would be $\sqrt{\frac{m}{k}}\frac{\pi}{2}$ I see my error. I confused $\sin^{-1}1$ with $\tan^{-1}1$ which was an earlier result.

Thank you,
kQ6Up