I am trying to work though an example on this topic in my book and have reached a point that I am not sure about. I was wondering if anyone could help me clear this up.(adsbygoogle = window.adsbygoogle || []).push({});

The equation of motion for a spring-mass system with no damping and a periodic external force is

[tex]mu'' + ku = Fcos\omega t[/tex]

The general solution to this is:

[tex] u = Acos\omega_0t + B sin\omega_0 t +\frac{F}{m(\omega_0^2+\omega^2)}cos\omega t[/tex]

If the mass is initially at rest, so that u(0)=0 and u'(0)=0, then the solution to this equation is

[tex]u=\frac{F}{m(\omega_0^2-\omega^2)}(cos\omega t -cos\omega_0 t)[/tex]

I have managed to follow it to here but I can not see how they have completed the next step:

"making use of the trigonometric identities for cos(A[itex]\pm[/itex]B) with [itex]A=(\omega_0+\omega)t/2[/itex] and [itex]B=(\omega_0-\omega)t/2[/itex] we can write the equation in the form":

[tex]u=\left[\frac{2F}{m(\omega_0^2-\omega^2)}sin\frac{(\omega_0-\omega)t}{2}\right]sin\frac{(\omega_0+\omega)t}{2}[/tex]

I can not see how the penultimate equation becomes the final equation here; can anyone tell me how this works?

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