Angular Frequency of Wave - General Physics - Solved Past Paper, Exams of Physics

Download Angular Frequency of Wave - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 8. (20 pts) A transverse sinusoidal wave on a string has a frequency of 5 Hz. The amplitude of the motion is 12 cm. It is traveling at 20 m/s to the left. At the origin at t = 0 s, the string is not displaced from equilibrium. a) (5 pts) What is the angular frequency of the wave? ω = 2πf = 10π Note that the units of angular frequency are radians/second, not Hz. b) (5 pts) Compute the wave number k. v = λf 20 = 5λ λ = 4 m Note that we are not talking about an electro-magnetic wave here, so the speed is not 3×108 m/s! Then, using the definition of k, we have k = 2π λ = 2π 4 = π 2 c) (5 pts) Write an expression for the wave function of this wave. y(x, t) = 0.12 sin (π 2 x + 10πt ) The phase is zero since sin(0) = 0 satisfies the indicated boundary condition. I suppose it is ok if the coefficient of the sin() function is 12 rather than 0.12, just remember that the units of the wave function are then cm rather than meters. d) (5 pts) Work out the maximum transverse speed of the string as the wave passes. Since the wave function of a transverse wave describes the transverse displacement of the medium as a function of both horizontal position and time, the transverse speed of the medium can be found by simply taking the rate of change of the transverse displacement (i.e. the wave function) with time: vy = ∂y ∂t vy = ∂ ∂t { 0.12 sin (π 2 x + 10πt )} vy = 1.2π cos (π 2 x + 10πt ) The maximum transverse velocity occurs when the cos() term has its maximum value: 1. The maxi- mum transverse velocity is therefore 1.2π = 3.8 m/s. The other possiblilty, if you tried the approach of memorizing hundreds of special equations rather than being able to apply a few fundamental concepts, is to recall equation (13.15) from the text, vy,max = ωA and just plug in the numbers.