Relation Between Spring Constant and Angular Velocity

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1. Nov 17, 2015

Simanto Rahman

I was going through Periodic Motion chapter of my book and came across an equation while defining the relation between Time Period of on oscillating particle and force constant.
k/m=w2

which was applied in, T = 2xpie/angular velocity

can anyone please help me define this equation. I can't seem to find any explanation for it in my text book.

2. Nov 17, 2015

stockzahn

ω = angular velocity in rad / s

2⋅π corresponds to one full rotation (2⋅π rad = 360°). If you want to know how many rotations are made in one second, you calculate ω / (2⋅π) = f with f = frequency in s-1. Now you calculated how many rotations per second correspond to your ω. The frequency is the inverse value of the time T, which is needed for one rotation (if something rotates two times per second, it needs 1/2 seconds for one rotation).

→ T = 1 / f = (2⋅π) / ω

3. Nov 17, 2015

Simanto Rahman

I understand the relation between angular velocity and Time period. What I don't understand is the relation between angular velocity and Spring Constant/Force constant (K) which is given as k/m= w2

4. Nov 17, 2015

stockzahn

F is the force to bring a spring-mass system in it's initial position. The (same) force in the spring accelerates the mass in the opposite direction:

Newton's 2nd law: F = - m⋅a = - m ⋅ ∂2x / ∂t2
Hooke's law: F = k ⋅ x

→ k ⋅ x = - m ⋅ ∂2x / ∂t2 → ∂2x / ∂t2 + k/m⋅x = 0

the solution of this DE can be obtained with x = A ⋅ cos (ω⋅t) → -A⋅ω2⋅cos(ω⋅t)+k/m⋅A⋅cos(ω⋅t) = 0 | / A , / cos(ω⋅t)

→ k/m = ω2

5. Nov 17, 2015

Simanto Rahman

But what I didn't get is how,
k ⋅ x = - m ⋅ ∂2x / ∂t2
becomes
2x / ∂t2 + k/m⋅x = 0

I see no way that displacement and mass can get into a multiplication format with each other. If we cross multiply it then we get, k.x/m + d2x/dt2 = 0

And I thought displacement of a particle in Periodic motion is,
x = Asin(w.t)
or, x = A.sin(w.t + δ) when δ is the initial phase.

6. Nov 17, 2015

Simanto Rahman

Alright I did the entire math on my notebook. Figured out your equation. its not k/(m.x) is (k/m).x
The use of brackets would've been nice but I got it. And I see now that you directly differentiated twice the value A.sin(w.t) for the last step. Thank you very much for your help!

7. Nov 17, 2015

stockzahn

There is no difference in these to equations, maybe I should have written it like this: ∂2x / ∂t2 + (k/m)⋅x = 0

At t = 0 the displacement is maximal and therefore A⋅cos (ω⋅t) = A⋅cos (0) = A. A sine curve and a cosine curve are the same, they are just shifted w.r.t. each other (by π/2).

A⋅sin(ω⋅t + π/2) = A⋅cos(ω⋅t)

But try to solve the DE with x = A⋅sin(ω⋅t + π/2) or x = A⋅sin(ω⋅t), you will obtain the same result.

8. Nov 17, 2015

Staff: Mentor

Perhaps the way that last equation is written is throwing you off. The second term is better written as (k/m)x.

(I see you've figured it out!)