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In forced oscillation the formula for the externally applied force is $F = \cos(\omega t)$ in almost every book except one book, which uses $F = \sin(\omega t)$.

If the equation for the position is $x = A\cdot\cos(\omega t+\phi)$ and velocity is the derivative of position, then the velocity of the oscillator should be proportional to $\sin(\omega t)$ (because the derivate of $\cos$ is $-\sin$), and hence match the function 'drawn' by the applied force which is actually '$\sin$' only in one book. But if it's '$\cos$' it just doesn't make any sense in terms of the resonance.

Which is correct? Are both of them correct? Why?

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  • $\begingroup$ I think you mean "then the velocity of the oscillator should be proportional to $\sin(\omega t +\phi)$." $\endgroup$ Feb 13, 2020 at 20:36
  • $\begingroup$ Do you know about Fourier transforms? $\endgroup$
    – DanielSank
    Feb 14, 2020 at 16:02

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In general there is a phase difference between the displacement, x, and the applied force, F. The phase difference depends on the frequency of F relative to the natural frequency of the oscillatory system. At resonance (or, more precisely, when the driving force frequency is the same as the system's undamped natural frequency) the displacement lags behind the driving force by $\tfrac{\pi}{2}$ (a quarter of a cycle).

It's usual to express both F and x as cosines or both as sines, so that the phase difference is simply the difference in the phase constants that are added to or subtracted from, $\omega t$. For example if $F=F_0 \cos (\omega t)$ and $x=x_0 \cos (\omega t +\phi)$, the displacement will be ahead of the driving force by a phase angle of $(\phi-0)=\phi$.

But it's perfectly possible to use $F=F_0 \sin (\omega t)$ for the force and $x=x_0 \cos (\omega t +\phi)$ for the displacement. Simply remember that $\sin (\omega t) =\cos (\omega t-\tfrac{\pi}{2})$. So in this case the displacement will be ahead of the driving force by a phase angle of $[\phi -(-\tfrac{\pi}{2})]=(\phi+\tfrac{\pi}{2})$. At resonance this phase angle is $-\tfrac{\pi}{2}$, so $(\phi+\tfrac{\pi}{2})=-\tfrac{\pi}{2}$, that is $\phi=-\pi$, which is indistinguishable from $\phi=\pi$.

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