Physics 325 Lecture 11 - Physics 325 Lecture 14...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Physics 325 Lecture 14 Superposition & Fourier Series We now investigate driven oscillations with non-sinusoidal driving forces. We’ll start with a simple superposition of two sinusoidal driving forces with different frequencies, and then show how to generalize this result to an arbitrary driving force. We take a driving force to be ( ) 11 2 cos cos Ft F t F t 2 ω =+ (14.1) and the equation of motion, which follows from Equation (13.2), is 2 12 01 2c o s c FF 2 o s x xx t mm t β ωω ++= + ±± ± (14.2) This equation, as before is linear and therefore the sum of solutions is also a solution. This inspires us to write two equations with solutions ( ) ( ) and x tx t : 2 1 0 1 1 2 2 22 0 2 F 2 o s o s x t m F x m t ± ± We can add these two equations to find () 2 2 0 1 2 2 2 cos cos dd x x x x t dt dt m m t ++ += + Therefore, () () x t + is a solution to Equation (14.2). We’ve demonstrated this for a superposition of two terms, but it obviously works for an arbitrary number of terms. Let () () () 1 cos N nn n n n F tF t F t F t = = = (14.3) Then for each n we can write 2 0 2 n nnn F c o s n x m t ± (14.4) and 1 N n n x = = t (14.5)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
satisfies the differential equation ( ) 2 0 2 Ft xxx m βω ++= ±± ± (14.6) This is the principle of superposition , and it works because the differential equation (14.6) is linear. In solving the problem of a sinusoidal driving force in Lecture 13, we have already found solutions to the differential equation (14.4). Therefore we can make the following statement: If the driving force F(t) has the form ( ) ( ) cos nn n F t n ω φ =− (14.7) and the oscillator system obeys Equation (14.6), then the solution for the resulting steady- state motion (long times) is () ( 2 2 22 0 1 cos 2 n pn F xt t m ) n n δ ωω β −+ (14.8) with 1 2 0 2 tan n n n 2 = (14.9) Fourier’s Theorem
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

Physics 325 Lecture 11 - Physics 325 Lecture 14...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online