# Derive time period for spring in SHM

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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.

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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#2

(Original post by

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.

**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.

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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#6

(Original post by

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

**robin.murphy1999**)how would you derive t=2Pi square root l/g

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#8

(Original post by

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

**robin.murphy1999**)how would you derive t=2Pi square root l/g

(Original post by

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

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

This is derived from Newtons laws.

Not sure how much calculus you use, but it is involved

(from resolving the vectors when the pendulum is in motion)

For small angles , we can use the approximation that

So using the acceleration of the pendulum, we get

Using differential equation form, we get

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

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

As such, we can say that for springs yes?

Comparing that to the pendulum equation, we see that and so for mass on a spring

so therefore the equation of a pendulum is

As required

EDIT: Just realised you're using sorry :P

Using this instead, you skip the differential equation:

solving for gives

Now as is displacement of the pendulum,

so

Therefore so you know where to go from this ()

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#9

(Original post by

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

This is derived from Newtons laws.

Not sure how much calculus you use, but it is involved

(from resolving the vectors when the pendulum is in motion)

For small angles , we can use the approximation that

So using the acceleration of the pendulum, we get

Using differential equation form, we get

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

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

As such, we can say that for springs yes?

Comparing that to the pendulum equation, we see that and so for mass on a spring

so therefore the equation of a pendulum is

As required

**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

This is derived from Newtons laws.

Not sure how much calculus you use, but it is involved

(from resolving the vectors when the pendulum is in motion)

For small angles , we can use the approximation that

So using the acceleration of the pendulum, we get

Using differential equation form, we get

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

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

As such, we can say that for springs yes?

Comparing that to the pendulum equation, we see that and so for mass on a spring

so therefore the equation of a pendulum is

As required

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#10

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

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