Assignment 4 PHSGCOR04T Sound Model Questions & Answers Phase and Group velocity Department of Physics, Hiralal Mazumdar Memorial College For Women, Dakshineswar Course Prepared by Parthapratim Pradhan 60 Lectures ≡ 4 Credits April 14, 2020 1. Define phase velocity and group velocity. We know that a traveling wave or a progressive wave propagated through a medium in a +ve x direction with a velocity v, amplitude a and wavelength λ is described by 2π y = a sin (vt − x) (1) λ If n be the frequency of the wave, then wave velocity v = nλ in this situation the traveling wave can be written as 2π x y = a sin (nλt − x)= a sin 2π(nt − ) (2) λ λ This can be rewritten as 2π y = a sin (2πnt − x) (3) λ = a sin (ωt − kx) (4) 2π 2π where ω = T = 2πn is called angular frequency of the wave and k = λ is called propagation constant or wave vector. This is simply a traveling wave equation. The Eq. (4) implies that the phase (ωt − kx) of the wave is constant. Thus ωt − kx = constant Differentiating with respect to t, we find dx dx ω ω − k = 0 ⇒ = dt dt k This velocity is called as phase velocity (vp). Thus it is defined as the ratio between the frequency and the wavelength of a monochromatic wave. dx ω 2πn λ vp = = = 2π = nλ = = v dt k λ T This means that for a monochromatic wave Phase velocity=Wave velocity 1 Where as the group velocity is defined as the ratio between the change in frequency and change in wavelength of a wavegroup or wave packet dω ∆ω ω1 − ω2 vg = = = dk ∆k k1 − k2 of two traveling wave described by the equations y1 = a sin (ω1 t − k1 x)and y2 = a sin (ω2 t − k2 x). Proof Let us consider two traveling waves of same amplitude and different frequency are superposed to each other then the resultant wave should be y = y1 + y2 = a [sin (ω1 t − k1 x) + sin (ω2 t − k2 x)] (5) Using the identity A + B A − B sin A + sin B = 2sin cos 2 2 The resultant displacement of the wave is thus ω1 t − k1 x + ω2 t − k2 x ω1 t − k1 x − ω2 t + k2 x y = 2a sin cos (6) 2 2 (ω1 + ω2)t − (k1 + k2)x (ω1 − ω2)t − (k1 − k2)x = 2a sin cos (7) 2 2 (8) Since ω1 − ω2 =∆ω and k1 − k2 =∆k are small in compared to ω1 and k1 respectively thus ω1 + ω2 ≈ 2ω1 and k1 + k2 ≈ 2k1 Therefore ∆ωt − ∆kx y = 2a sin (ω1t − k1x) cos (9) 2 This equation indicates that a wave of angular frequency ω1 and wave number k that has ∆ω ∆k superposed upon it a modulation of angular frequency 2 and of wave number 2 . Thus the phase velocity of the resultant wave is ω1 vp = k1 and the group velocity of the wave group is ∆ω dω 2 ω1 − ω2 v = = = g ∆k − dk 2 k1 k2 2. Find the relation between phase velocity and group velocity. We know ω v = ⇒ ω = v k p k p and dω d dv v = = (v k)= v + k p (10) g dk dk p p dk dvp dλ = vp + k (11) dλ dk 2 Again 2π 2π k = ⇒ λ = λ k After differentiation, we get dλ 2π = − dk k2 Putting these values in Eq. (11), we find dvp 2π vg = vp + k − (12) dλ k2 2π dvp = vp + − (13) k dλ dvp = vp − λ (14) dλ This is the relation between group velocity and phase velocity. dvp Note that in case of dispersive medium, the velocity increases with wavelength thus dλ is positive always. Therefore vg < vp. 3.