# Formula for the harmonic oscillator is f"(x)+W^2 * X(t)=0

Chiara
can anyboy show to me why the formula for the harmonic oscillator is f"(x)+W^2 * X(t)=0. Please I spent a whole afternoon trying to figure it out and I just wasted my time.
Thanks

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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,

$$x(t) = A \sin(\omega t+\phi)$$

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

$$\ddot{x}(t) + \omega^2 x(t) = 0$$

Hopefully i can get this right.

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.