Question

(3 pts) Mechanical energy of a segment (of length dx) of string is given by ME ufdx 1. segment However, this can vary with segment location (sometimes the maximum value and sometimes zero). A more useful expression is the mechanical energy over one wavelength. To find the energy for a wave length, you add up the ME of each segment for one wavelength: ME one wavelength ME segmentdx 0 You can evaluate this at any time you choose. So, using this expression, show that the mechanical energy of one wavelength for a wave with a wave function given by y(x, t) Asin(kx-wt) is equal to 1A2ω2λ. 2

Answer 1

ME_segment = T/V^2_p V_t^2 dx V_ = squareroot T/mu implies V^2_p = T/mu and y = A sin (kx - wt) implies V_t = dy/dt = -A w cos (kx -wt) implies V^2_t = A^2 W^2 cos^2 (kx - wt) therefore ME_segment = T/T/mu A^2 W^2 cos^2 (kx - wt) dx = mu A^2 W^2 cos^2 (kx - wt)dx ME_segment = integral^lambda_0 ME_segment dx = integral^lambda_0 mu A62 W^2 cos^2 (kx - wt) dx Let t = 0 then ME_one wavelength = mu A62 W^2 integral^lambda_0 Ws^2 (kx) dx = mu A^2 W^2 /2 integral^lambda_0 (1- cos Q kx))dx \r\n (therefore cos^2 theta = 1 + cos 2 theta /2) implies ME_one wavelength = mu A^2 W^2/2 {integral^lambda_0 dx + integral^lambda_0 cos (2kx) dx mu A^2 W^2 /2 {[x]^lambda_0 + [sin 2kx/2k]^lambda_0 } = mu A^2 W^2/2