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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
Ambitwistor
Nov19-03, 03:12 PM
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
Beer-monster
Nov20-03, 05:42 AM
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.
[:D]
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