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

AI Thread Summary
The harmonic oscillator's equation f"(x) + W^2 * X(t) = 0 derives from Newton's second law applied to a mass-spring system. The forces acting on the mass include the inertial force (F1 = ma) and the spring's restoring force (F2 = -kx). Setting these forces equal results in the equation ma + kx = 0, which simplifies to mx"(t) + kx = 0. Dividing by m leads to the standard form x"(t) + (k/m)x = 0, where the angular frequency w is defined as w = sqrt(k/m). Substituting w^2 into the equation confirms the harmonic oscillator's standard form, demonstrating the relationship between mass, spring constant, and oscillation.
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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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 that's understandable as I never tried writing such an equation in pure text before. Hope it works and helps.
 
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