(f) The speed of the wave is 2510rad s40m s.62.8rad/mvTkωλ====70. We write the expression for the displacement in the form y(x, t) = ymsin(kx– t).(a) The amplitude is ym= 2.0 cm = 0.020 m, as given in the problem. (b) The angular wave number kis k= 2π/λ= 2π/(0.10 m) = 63 m–1(c) The angular frequency is = 2πf= 2π(400 Hz) = 2510 rad/s = 2.5×103rad/s. (d) A minus sign is used before the tterm in the argument of the sine function because
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.