- #1

- 7

- 2

k/m=w

^{2}

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.

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- Thread starter Simanto Rahman
- Start date

- #1

- 7

- 2

k/m=w

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

stockzahn

Homework Helper

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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

→ T = 1 / f = (2⋅π) / ω

- #3

- 7

- 2

- #4

stockzahn

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Newton's 2nd law: F = - m⋅a = - m ⋅ ∂

Hooke's law: F = k ⋅ x

→ k ⋅ x = - m ⋅ ∂

the solution of this DE can be obtained with x = A ⋅ cos (ω⋅t) → -A⋅ω

→ k/m = ω

- #5

- 7

- 2

Newton's 2nd law: F = - m⋅a = - m ⋅ ∂^{2}x / ∂t^{2}

Hooke's law: F = k ⋅ x

→ k ⋅ x = - m ⋅ ∂^{2}x / ∂t^{2}→ ∂^{2}x / ∂t^{2}+ k/m⋅x = 0

the solution of this DE can be obtained with x = A ⋅ cos (ω⋅t) → -A⋅ω^{2}⋅cos(ω⋅t)+k/m⋅A⋅cos(ω⋅t) = 0 | / A , / cos(ω⋅t)

→ k/m = ω^{2}

But what I didn't get is how,

k ⋅ x = - m ⋅ ∂

becomes

∂

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 + d

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

- 7

- 2

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

stockzahn

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∂^{2}x / ∂t^{2}+ k/m⋅x = 0

k.x/m + d^{2}x/dt^{2}= 0

There is no difference in these to equations, maybe I should have written it like this: ∂

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.

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

Doc Al

Mentor

- 45,254

- 1,615

Perhaps the way that last equation is written is throwing you off. The second term is better written as (k/m)x.But what I didn't get is how,

k ⋅ x = - m ⋅ ∂^{2}x / ∂t^{2}

becomes

∂^{2}x / ∂t^{2}+ k/m⋅x = 0

(I see you've figured it out!)

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