- #1

Thanks

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- Thread starter Chiara
- Start date

- #1

Thanks

- #2

Ambitwistor

- 841

- 1

Originally posted by Chiara

can anyboy show to me why the formula for the harmonic oscillator is f"(x)+W^2 *X(t)=0.

Do you mean x''(t), not f''(x)?

I'm not sure where you're starting from. Are you starting from a harmonic oscillator as something whose motion is defined by,

[tex]x(t) = A \sin(\omega t+\phi)[/tex]

? If so, if you just calculate its second time derivative [tex]\ddot{x}(t)[/tex], you can see immediately that it satisfies the equation,

[tex]\ddot{x}(t) + \omega^2 x(t) = 0[/tex]

- #3

Beer-monster

- 296

- 0

Using the example of a mass and spring oscillator. By Newton's £rd Law the forces on the spring are equal. The forces on the spring are the force on the mass (F1=ma) and the resistive force on the spring

(F2+-kx, negative because it is resitive and opposite to F1=ma)

Therefore

F1=F2

ma=-kx

ma+kx=0

By definition a= x"(t)

So mx"(t)+ kx=0

Dividing by m gives x"(t)+(k/m)x=0

The standard equation for the period of oscillation (T) of a mass spring

pendulum is

T = (2(Pi))sqrt(m/K)

omega= w = (2(Pi))/T = 2(pi)/((2(Pi))sqrt(m/K))

The 2(pi) should cancel and leave you with w = sqrt(K/m)

Squaring both sides gives w^2=(K/m)

Substitute this into the above equation and get

x"(t) + (w^2)x = 0

I hope thats understandable as I never tried writing such an equation in pure text before. Hope it works and helps.

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