# Derive time period for spring in SHM

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Thread starter 7 years ago
#1
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.
1
7 years ago
#2
(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 where is the displacement from the origin. We usually write the associated equation in the form , because then after solving, we find that the period is given by .

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)
0
Thread starter 7 years ago
#3
Ah right:

sub back in:
0
Thread starter 7 years ago
#4
Ah right:

sub back in:

I didn't think to use the acceleration equation for SHM.
Thanks for the help!
0
4 years ago
#5
how would you derive t=2Pi square root l/g
0
4 years ago
#6
(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
0
4 years ago
#7
so how do you get t = 2Pi (l/g)^1/2
0
4 years ago
#8
(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

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 ()
0
4 years ago
#9
(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

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
Brilliant, thank you
0
3 years ago
#10
Show that for a body vibration simple harmonically the time period ia given by T=2π√(L/g)
0
2 years ago
#11
Thankssssss
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