Derive time period for spring in SHM

TheSK00T3R

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Derive T=2Pi(m/k)^1/2

I've got as far as T=2Pi/w => 1/w=(m/k)^1/2 => w^2=k/m
I don't even know what k/m is.
I've tried integrating trig graphs, rearranging spring constant equations, all to no avail.
If anyone could give me a pointer that would be much appreciated.

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atsruser

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(Original post by TheSK00T3R)
Derive T=2Pi(m/k)^1/2

I've got as far as T=2Pi/w => 1/w=(m/k)^1/2 => w^2=k/m
I don't even know what k/m is.
I've tried integrating trig graphs, rearranging spring constant equations, all to no avail.
If anyone could give me a pointer that would be much appreciated.

The defining feature of SHM is that $a \propto -x$ where

is the displacement from the origin. We usually write the associated equation in the form $a=-\omega^2 x$ , because then after solving, we find that the period is given by $T=\frac{2\pi}{\omega}$ .

For a mass on a spring, the restoing force is given by

where

is the spring constant.

Can you finish it from here? (Hint: rearrange and equate coefficients)

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TheSK00T3R

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Ah right:
$ma=-kx a=\dfrac{-kx}{m} a=-\omega^2x \dfrac{-kx}{m}=-\omega^{2}x => \dfrac{-kx}{mx}=-\omega^{2}=>w=\sqrt{\dfrac{k}{m} }$
sub $\omega$ back in:
$T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m }}}={2\pi}\sqrt{\dfrac{m}{k}}$

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TheSK00T3R

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Ah right:
$ma=-kx a=\dfrac{-kx}{m} a=-\omega^2x \dfrac{-kx}{m}=-\omega^{2}x => \dfrac{-kx}{mx}=-\omega^{2}=>w=\sqrt{\dfrac{k}{m} }$
sub $\omega$ back in:
$T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m }}}={2\pi}\sqrt{\dfrac{m}{k}}$
I didn't think to use the acceleration equation for SHM.
Thanks for the help!

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robin.murphy1999

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how would you derive t=2Pi square root l/g

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TheFarmerLad

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(Original post by robin.murphy1999)
how would you derive t=2Pi square root l/g

This ^. I can prove the period equation for springs (the equation the OP stated) but not for a pendulum

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robin.murphy1999

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so how do you get t = 2Pi (l/g)^1/2

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The-Spartan

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(Original post by robin.murphy1999)
how would you derive t=2Pi square root l/g

(Original post by TheFarmerLad)
This ^. I can prove the period equation for springs (the equation the OP stated) but not for a pendulum

You can kind of think of it as a simple harmonic motion equation because the angular displacement of a pendulum is given as

$\theta=\theta_{max} \sin{\sqrt{\frac{g}{l}}T}$

This is derived from Newtons laws.
Not sure how much calculus you use, but it is involved

$F_{net}=-mg\sin{\theta}$ (from resolving the vectors when the pendulum is in motion)

For small angles $\theta$ , we can use the approximation that $\sin{\theta} \approx \theta$

So using the acceleration of the pendulum, we get

$F_{net}=-mg\theta=maL$

Using differential equation form, we get

$\frac{d^2\theta}{dt^2} + \frac{g\theta}{L}=0$

Solving this equation gets you to the equation at the start.

Anyway, this equation can be likened to a mass on a spring

$x=A\sin{\sqrt{\frac{k}{m}}T}$

As such, we can say that $\omega = \sqrt{\frac{k}{m}}$ for springs yes?

Comparing that to the pendulum equation, we see that $\omega = \sqrt{\frac{g}{L}}$ and so for mass on a spring

$T=2\pi\sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1}{\omega}}$ so therefore the equation of a pendulum is

$T=2\pi \sqrt{\frac{L}{g}}$
As required

EDIT: Just realised you're using $a=-\omega^2x$ sorry :P
Using this instead, you skip the differential equation:
$F_{net}=maL=mg\theta$ solving for

gives $a=-\frac{g\theta}{L}$

Now as $\theta$ is displacement of the pendulum, $\theta=x$

so $-\frac{gx}{L}=-\omega^2x$
Therefore $\omega = \sqrt{\frac{g}{L}}$ so you know where to go from this ( $\omega = 2\pi f$ )

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TheFarmerLad

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(Original post by The-Spartan)
You can kind of think of it as a simple harmonic motion equation because the angular displacement of a pendulum is given as

$\theta=\theta_{max} \sin{\sqrt{\frac{g}{l}}T}$

This is derived from Newtons laws.
Not sure how much calculus you use, but it is involved

$F_{net}=-mg\sin{\theta}$ (from resolving the vectors when the pendulum is in motion)

For small angles $\theta$ , we can use the approximation that $\sin{\theta} \approx \theta$

So using the acceleration of the pendulum, we get

$F_{net}=-mg\theta=maL$

Using differential equation form, we get

$\frac{d^2\theta}{dt^2} + \frac{g\theta}{L}=0$

Solving this equation gets you to the equation at the start.

Anyway, this equation can be likened to a mass on a spring

$x=A\sin{\sqrt{\frac{k}{m}}T}$

As such, we can say that $\omega = \sqrt{\frac{k}{m}}$ for springs yes?

Comparing that to the pendulum equation, we see that $\omega = \sqrt{\frac{g}{L}}$ and so for mass on a spring

$T=2\pi\sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1}{\omega}}$ so therefore the equation of a pendulum is

$T=2\pi \sqrt{\frac{L}{g}}$
As required

Brilliant, thank you

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Shahosh

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Show that for a body vibration simple harmonically the time period ia given by T=2π√(L/g)

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Khing

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Thankssssss

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