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Time to Max Compression of a Spring

  1. May 16, 2017 #1
    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. jcsd
  3. May 16, 2017 #2

    kuruman

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    Science Advisor
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    Almost correct. Have you studied simple harmonic motion? If so, what fraction of the period is the time from equilibrium to maximum displacement?
     
  4. May 16, 2017 #3
    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
     
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