- #1
jellicorse
- 40
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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.
The equation of motion for a spring-mass system with no damping and a periodic external force is
mu'' + ku = Fcos\omega t
The general solution to this is:
u = Acos\omega_0t + B sin\omega_0 t +\frac{F}{m(\omega_0^2+\omega^2)}cos\omega t
If the mass is initially at rest, so that u(0)=0 and u'(0)=0, then the solution to this equation is
u=\frac{F}{m(\omega_0^2-\omega^2)}(cos\omega t -cos\omega_0 t)
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\pmB) with A=(\omega_0+\omega)t/2 and B=(\omega_0-\omega)t/2 we can write the equation in the form":
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}
I can not see how the penultimate equation becomes the final equation here; can anyone tell me how this works?
The equation of motion for a spring-mass system with no damping and a periodic external force is
mu'' + ku = Fcos\omega t
The general solution to this is:
u = Acos\omega_0t + B sin\omega_0 t +\frac{F}{m(\omega_0^2+\omega^2)}cos\omega t
If the mass is initially at rest, so that u(0)=0 and u'(0)=0, then the solution to this equation is
u=\frac{F}{m(\omega_0^2-\omega^2)}(cos\omega t -cos\omega_0 t)
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\pmB) with A=(\omega_0+\omega)t/2 and B=(\omega_0-\omega)t/2 we can write the equation in the form":
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}
I can not see how the penultimate equation becomes the final equation here; can anyone tell me how this works?
