Let us envision three frames of reference I, I’ and I” in standard configuration with I’ moving with velocity V relative to I and I” moving with velocity u relative to I and with velocity u’ relative to I’. Let O, O’ and O” be the spatial origins of the frames respectively; O, O’ and O” coinciding at the origin of time in the three frames (t=t’=0). Accept the following notations g(V)=1/sqr(1-VV/cc); g(u)=1/sqr(1-uu/cc) and g(u’)=1/sqr(1-u’u’/cc). Let I” be the rest frame of a rod located along the permanently overlapped axes x,x’x” where its length is L0. Observers from I detect simultaneously the space coordinates of the ends of the rod, subtract them obtaining that its length is L related to L0 by L=L0/g(u). (1) Following the same procedure observers from I’ obtain for the length of the moving rod L’=L0/g(u’). (2) Combining (1) and (2) we obtain that L and L’ are related by L=L’g(u’)/g(u). (3) Expressing the right side of (3) as a function of u’ via the addition law of parallel speeds u=(u’+V)/(1+Vu’/cc) (4) the result is L=L’/[g(V)(1+Vu’/cc)]. (5) I underline that (4) could be derived without invoking the Lorentz transformations. Introducing the physical quantities k=1/L and k’=1/L’ we obtain that they transform as k=g(V)k’(1+Vu’/cc). (6) If the rod is generated by a particle moving with speed u for a proper time interval t(0) i.e. L=ut(0) (7) in I and L’=u’t’(0) (8) then (6) becomes k=g(V)[k’+V/t’(0)cc]=g(V)[k’+Vf’/cc] (9) where f’=1/t’(0) has the physical meaning of frequency. If the rods are generated by light signals then L=ct(0) and L’=ct’(0) (6) becoming k=k’sqr[(1+V/c)/(1-V/c)] . (10) We obtain the transformation equations for the frequencies f and f’ starting with f=ku and f’=k’u’. Combining them and taking into account (4) the result is f=g(V)f’(1+V/u’)=g(V)(f’+Vk’) (11) which becomes in the case of rods generated by light signals f=f’sqr[(1+V/c)/(1-V/c)]. (12) Would you accept all that as a introduction in special relativity in the acoustic and in the electromagnetic plane waves?