Derive time period for spring in SHM

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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 x 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 F = ma = -kx where k 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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#5
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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 a 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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