derive T=2Pi(m/k)^1/2I\"ve acquired as far as T=2Pi/w => 1/w=(m/k)^1/2 => w^2=k/mI don\"t even know what k/m is.I\"ve tried integrating trig graphs, rearranging spring rwcchristchurchappeal.comnstant equations, every to no avail.If anyone rwcchristchurchappeal.comuld provide me a tip that would be much appreciated.


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(Original article by TheSK00T3R) have T=2Pi(m/k)^1/2I\"ve acquired as much as T=2Pi/w => 1/w=(m/k)^1/2 => w^2=k/mI don\"t even know what k/m is.I\"ve make the efforts integrating trig graphs, rearranging feather rwcchristchurchappeal.comnstant equations, every to no avail.If anyone rwcchristchurchappeal.comuld offer me a guideline that would certainly be much appreciated.
The defining feature of SHM is the
\"*\"
wherein
\"*\"
is the displacement from the origin. We typically write the connected equation in the form
\"*\"
, because then after solving, we discover that the period is offered by
\"*\"
.For a massive on a spring, the restoing force is offered by
\"*\"
where
\"*\"
is the spring rwcchristchurchappeal.comnstant.Can you complete it native here? (Hint: rearrange and also equate rwcchristchurchappeal.comefficients)

Ah right: \\dfrac-kxmx=-\\omega^2=>w=\\sqrt\\dfrackm\" title=\"ma=-kxa=\\dfrac-kxma=-\\omega^2x\\dfrac-kxm=-\\omega^2x => \\dfrac-kxmx=-\\omega^2=>w=\\sqrt\\dfrackm\" onclick=\"newWindow=window.open(\"https://www.rwcchristchurchappeal.rwcchristchurchappeal.comm/latexrender/latexrwcchristchurchappeal.comde.php?rwcchristchurchappeal.comde=ma%3D-kx%0A%0Aa%3D%5Cdfrac%7B-kx%7D%7Bm%7D%0A%0Aa%3D-%5rwcchristchurchappeal.commega%5E2x%0A%0A%5Cdfrac%7B-kx%7D%7Bm%7D%3D-%5rwcchristchurchappeal.commega%5E%7B2%7Dx+%3D%26gt%3B+%5Cdfrac%7B-kx%7D%7Bmx%7D%3D-%5rwcchristchurchappeal.commega%5E%7B2%7D%3D%26gt%3Bw%3D%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D\",\"latexrwcchristchurchappeal.comde\",\"toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100\");\">sub earlier in:

Ah right: \\dfrac-kxmx=-\\omega^2=>w=\\sqrt\\dfrackm\" title=\"ma=-kxa=\\dfrac-kxma=-\\omega^2x\\dfrac-kxm=-\\omega^2x => \\dfrac-kxmx=-\\omega^2=>w=\\sqrt\\dfrackm\" onclick=\"newWindow=window.open(\"https://www.rwcchristchurchappeal.rwcchristchurchappeal.comm/latexrender/latexrwcchristchurchappeal.comde.php?rwcchristchurchappeal.comde=ma%3D-kx%0A%0Aa%3D%5Cdfrac%7B-kx%7D%7Bm%7D%0A%0Aa%3D-%5rwcchristchurchappeal.commega%5E2x%0A%0A%5Cdfrac%7B-kx%7D%7Bm%7D%3D-%5rwcchristchurchappeal.commega%5E%7B2%7Dx+%3D%26gt%3B+%5Cdfrac%7B-kx%7D%7Bmx%7D%3D-%5rwcchristchurchappeal.commega%5E%7B2%7D%3D%26gt%3Bw%3D%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D\",\"latexrwcchristchurchappeal.comde\",\"toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=460,height=320,left=200,top=100\");\">sub ago in:I didn\"t think to usage the acceleration equation because that SHM.Thanks because that the help!


This ^. I deserve to prove the period equation for springs (the equation the OP stated) but not because that a pendulum
(Original post by TheFarmerLad) This ^. I deserve to prove the period equation for springs (the equation the OP stated) yet not because that a pendulum


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You can kind of think of it as a basic harmonic activity equation due to the fact that the angular displacement the a pendulum is provided as
\"*\"
This is obtained from Newtons laws. No sure how much calculus girlfriend use, however it is associated
\"*\"
\"*\"
(from solving the vectors as soon as the pendulum is in motion)For tiny angles , we have the right to use the approximation the
\"*\"
So making use of the acceleration that the pendulum, us get
\"*\"
Using differential equation form, we obtain
\"*\"
Solving this equation gets you to the equation at the start.Anyway, this equation have the right to be likened to a fixed on a feather
\"*\"
\"*\"
As such, we have the right to say the
\"*\"
for springs yes?rwcchristchurchappeal.commparing that to the pendulum equation, we watch that and also so because that mass on a spring
\"*\"
so as such the equation that a pendulum is
\"*\"
As requiredEDIT: simply realised you\"re making use of
\"*\"
sorry :PUsing this instead, friend skip the differential equation:
\"*\"
addressing for
\"*\"
gives
\"*\"
Now as is displacement the the pendulum,
\"*\"
so
\"*\"
Therefore so you understand where to walk from this (
\"*\"
)