Sum of Normal Modes on a Vibrating String

1 reply · 2K views · last reply
MyNewPony
Messages
30
Reaction score
0

Homework Statement



In textbooks, I often see the sum of the first two normal modes given in the equation attached (on the right). I'm wondering how they arrive at that equation based on the general formula (on the left).

I tried subbing in n= 1 and 2 in the general formula, but I'm not sure where to go from there. Where does the phase angle in the first normal mode go? Why is the amplitude of the second normal mode 1/2?
 
Attachments
  • Equation for the displacement u(x,t) as an infinite sum over n of A_n cos(omega_n t + phi_n) sin(n pi x / l), equal to the sum of u_n(x,t).
    Normal Modes.jpg
    4.4 KB · Views: 501
  • Equation u(x,t) = cos(ω1 t) sin(πx/l) + (1/2) cos(ω2 t + π/2) sin(2πx/l), the sum of two normal modes of a vibrating string.
    First Two Normal Modes.jpg
    3.8 KB · Views: 559
The equation on the right is not general, but depends specifically on the initial conditions. Eg., the string's displacement and velocity at t=0.

The fact that the phase angles of the 2 modes are zero and π/2, or that the amplitudes are 1 and 1/2, is not a generally true result.