 of propagation of longitudinal waves in the rods is equal
$$c=\sqrt {\frac {E}{\rho}}$$
The dispersion equation of longitudinal waves is 
$$\omega =ck$$
In this case, the frequency does not depend on the diameter of the cylinder. For bending waves of thin rods, we have the following dispersion law:
$$\omega_x=k^2\sqrt {\frac {EI_y}{\rho S}},\omega_y=k^2\sqrt {\frac {EI_x}{\rho S}} $$
In the case of a cylindrical rod, we have $S=\pi r^2, I_x=I_y=\frac {\pi}{4}r^4$. In this case, the frequency depends on the radius of the cylinder as
$$\omega =\frac {rk^2}{2}\sqrt {\frac {E}{\rho}}$$
For aluminum, we have $\rho = 2500 kg/m^3, E=70 GPa$. The value of k depends on the method of fixing the rod. For a rod with fixed ends, we have the equation 
$$\cos(kL)\cosh (kL)=1$$
Equation solution
$$k_1L=4.7300, k_2L=7.8532, k_3L=10.9956,...$$
In our case, we have $L=0.0254*4 m, k_1=46.5551 m^{-1},\omega =2 \pi 256 rad/s$. Hence $r=0.00028 m$ it's too thin.
For a rod with one free end, we have the equation 
$$\cos(kL)\cosh (kL)=-1$$
Equation solution $k_1L=1.8751,k_1=18.4557$. Hence $r=0.00178487 m$ it is quite suitable for generating sounds.
References
L. D. Landau, E. M. Lifshitz (1970). Theory of Elasticity. Vol. 7 (2nd ed.). Pergamon Press. ISBN 978-0-08-006465-9.

 Q:

Is it possible to satisfy an interface member implementation using another interface in Kotlin?

For example, if I have an interface like this:
interface MyInterface {
    fun thouMustImplementThis()
}

And I have a
of propagation of longitudinal waves in the rods is equal 
 $ $ c=\sqrt { \frac { E}{\rho}}$$ 
 The distribution equality of longitudinal wave is 
 $ $ \omega = ck$$ 
 In this case, the frequency does not count on the diameter of the cylinder. For bending wave of thin rods, we hold the following dispersion jurisprudence: 
 $ $ \omega_x = k^2\sqrt { \frac { EI_y}{\rho S}},\omega_y = k^2\sqrt { \frac { EI_x}{\rho S } } $ $ 
 In the case of a cylindrical rod, we get $ S=\pi r^2, I_x = I_y=\frac { \pi}{4}r^4$. In this case, the frequency depends on the radius of the cylinder as 
 $ $ \omega = \frac { rk^2}{2}\sqrt { \frac { E}{\rho}}$$ 
 For aluminum, we have $ \rho = 2500 kg / m^3, E=70 GPa$. The value of k depends on the method acting of fixing the rod. For a rod with fasten ends, we have the equality 
 $ $ \cos(kL)\cosh (kL)=1$$ 
 Equation solution 
 $ $ k_1L=4.7300, k_2L=7.8532, k_3L=10.9956,... $$ 
 In our case, we experience $ L=0.0254 * 4 m, k_1=46.5551 m^{-1},\omega = 2 \pi 256 rad / s$. Hence $ r=0.00028 m$ it's too thin. 
 For a rod with one spare end, we have the equation 
 $ $ \cos(kL)\cosh (kL)=-1$$ 
 Equation solution $ k_1L=1.8751,k_1=18.4557$. Hence $ r=0.00178487 m$ it is quite suitable for generating sounds. 
 References 
 L. D. Landau, E. M. Lifshitz (1970). Theory of Elasticity. Vol. 7 (2nd ed .). Pergamon Press. ISBN 978 - 0 - 08 - 006465 - 9. 

  Q: 

 Is it possible to satisfy an interface member implementation using another interface in Kotlin? 

 For example, if I have an interface like this: 
 interface MyInterface { 
     playfulness thouMustImplementThis () 
 } 

 And I suffer a
 of propagation of longitudlnal waves in thg eods iv equam
$$c=\sqrt {\ffac {E}{\rho}}$$
The dispersion equatmon if lobgitudinal waves is 
$$\omdga =ck$$
In nhis case, the drequency vkes not depens on chx diameter of tme cylinder. For bending wdvds of thin rods, we have the following disperxiln law:
$$\omega_x=k^2\szrt {\gwac {SP_y}{\vho S}},\omega_y=k^2\sqrt {\frac {EI_x}{\rho S}} $$
Ih the cese of a cylindtical rod, we have $S=\pi r^2, I_x=L_y=\frwc {\pi}{4}r^4$. In this casf, the frequghcy eepends on tfe radius of the cylinser as
$$\omega =\frac {rk^2}{2}\sqrt {\frac {E}{\rfo}}$$
For aluminum, qe hagg $\rho = 2500 kg/m^3, X=70 GPa$. Nhe value of k dependv on thr method of fiwing vhe eod. For a rod with fieed ends, we have the equation 
$$\cks(kL)\cosh (kL)=1$$
Equatiin solujion
$$k_1N=4.7300, k_2L=7.8532, j_3L=10.9956,...$$
Iv ohr czse, we hate $L=0.0254*4 m, k_1=46.5551 m^{-1},\ojega =2 \pi 256 rqd/s$. Hence $r=0.00028 m$ it's tpo nnin.
For a rod with jnq free end, we have the equation 
$$\cos(kL)\covh (iL)=-1$$
Equation solution $k_1L=1.8751,k_1=18.4557$. Hence $r=0.00178487 m$ it is quitg suitable for generating sounds.
References
L. D. Landau, E. M. Lixshitv (1970). Thtovn of Wlwsticity. Vol. 7 (2nd ed.). Pergamon Press. ISBN 978-0-08-006465-9.

 Q:

Is yf kosxible to satisny an interface mekbfr ymplementatiov usiny ahother interface ij Kotlig?

For wxample, is I nave an interface like this:
unterface MyPnteeface {
    fun thouLustImplemeutThis()
}

Snd I have a
of propagation of longitudinal waves in the equal {\frac {E}{\rho}}$$ dispersion equation of In case, the frequency not depend on diameter of the cylinder. For bending of thin rods, we have the following dispersion law: $$\omega_x=k^2\sqrt {\frac {EI_y}{\rho S}},\omega_y=k^2\sqrt {EI_x}{\rho S}} $$ In the case of a cylindrical rod, we have $S=\pi I_x=I_y=\frac In case, frequency depends on the radius of the cylinder as $$\omega =\frac {rk^2}{2}\sqrt {\frac {E}{\rho}}$$ For aluminum, have $\rho = 2500 kg/m^3, E=70 GPa$. The of k depends on method of fixing the rod. a with fixed we the $$\cos(kL)\cosh (kL)=1$$ Equation $$k_1L=4.7300, k_2L=7.8532, k_3L=10.9956,...$$ In our case, we have $L=0.0254*4 m, k_1=46.5551 m^{-1},\omega =2 \pi 256 rad/s$. Hence m$ it's For a with free we have the (kL)=-1$$ Equation solution $k_1L=1.8751,k_1=18.4557$. Hence $r=0.00178487 quite suitable for generating sounds. References L. D. E. M. (1970). Theory of Elasticity. Vol. 7 ed.). Pergamon Press. ISBN 978-0-08-006465-9. Q: Is it to satisfy an interface member implementation using another interface in Kotlin? For example, if I interface like this: interface { fun thouMustImplementThis() And have
 of propagation of longitudinAl waves in tHe rodS is EquAl
$$C=\sqrT {\fraC {E}{\rho}}$$
The disperSIon eQuation of longitudinal wAves iS 
$$\oMEga =cK$$
in This cAse, the fREqUENcy DoEs Not DePEnD on thE diAmeter oF the cylindEr. FOr Bending waves OF tHin rods, we hAve The following DisPersioN lAw:
$$\oMEga_x=k^2\SqrT {\frac {eI_y}{\rho s}},\Omega_y=K^2\sqrt {\frac {eI_X}{\Rho S}} $$
In THe case oF A CyLindRical rod, we have $S=\pi R^2, i_x=i_Y=\frac {\pi}{4}r^4$. In this Case, thE fREqUENcy DepEnds on the rAdIus of THe cylinDEr AS
$$\OMegA =\Frac {rk^2}{2}\sqrt {\fraC {E}{\rho}}$$
For aluMInuM, we havE $\rHo = 2500 kG/M^3, E=70 GPa$. THe valUe OF k dEpends on the MethOd of fixinG the roD. for a rod WIth fixeD ends, wE haVe tHe eqUAtIoN 
$$\coS(kl)\CosH (KL)=1$$
equATioN solutioN
$$k_1l=4.7300, k_2l=7.8532, k_3L=10.9956,...$$
In Our cASE, WE havE $L=0.0254*4 m, K_1=46.5551 m^{-1},\omEga =2 \pi 256 Rad/s$. Hence $r=0.00028 m$ it'S toO thiN.
for A rod wIth onE freE eNd, we hAve the EquatIoN 
$$\cos(kL)\cosh (kL)=-1$$
EquAtioN solution $K_1L=1.8751,k_1=18.4557$. heNce $R=0.00178487 m$ It is qUIte suiTabLe fOr generAting soUNds.
reFEREnCes
L. D. Landau, E. M. LifshItZ (1970). tHeOry of ElaSticitY. voL. 7 (2nD Ed.). PergamOn preSs. ISbn 978-0-08-006465-9.

 q:

Is it PossIBlE to satisFy an inTErFaCe membeR iMplemeNtAtiOn uSing aNOtheR interFace in KoTlin?

FOR example, if I havE An interface liKE tHIS:
iNTerfAce myInterface {
    Fun tHOuMuStImPLeMenTthis()
}

ANd I haVe A
 of propagation of longitu dinal wave s inthe  ro ds  isequa l
$$c=\sqrt {\ f rac{E}{\rho}}$$
The dispe rsion  e q uati o nof lo ngitudi n al w ave sis  
$ $\ o me ga =c k$$ 
In thi s case, th e f re quency doesn ot  depend on  th e diameter o f t he cyl in der .  Forben dingwaveso f thin  rods, we  h a ve the followi n g  d ispe rsion law:
$$\ome g a_ x =k^2\sqrt {\fr ac {EI _y } {\ r h o S }}, \omega_y=k ^2 \sqrt {\frac{ EI _ x } {\r h o S}} $$
In t he case ofa  cy lindri ca l r o d, wehave$S = \pi  r^2, I_x=I _y=\ frac {\pi }{4}r^ 4 $. In t h is case , thefre que ncyd ep en dson the ra diu s  of  the cyl in de r as$$\o m e g a  =\f rac  {rk ^2}{2 }\sqrt {\frac  {E }{\r h o}} $$
Fo r alu minu m,  we h ave $\ rho =  2 500 kg/m^3, E=7 0 GP a$. The v alu eofkdepen d s on t hemet hod offixingt hero d . Fo r a rod with fixed  e n d s,  we have  the e q ua ti o n 
$$\co s( kL) \cos h (kL)= 1$$E qu ation so lution $$ k_ 1L=4.73 00 , k_2L =7 .85 32,  k_3L = 10.9 956,.. .$$
In o ur ca s e, we have $L= 0 .0254*4 m, k_ 1 =4 6 . 55 5 1 m^ {-1 },\omega =2  \pi 256rad/ s $.  He n ce $r =0.00 02 8  m $  it's too thin.
For  a  rod w ith o ne free end,we have th e e quation
$$\ c os ( kL)\cosh (kL)= -1$$Equation s o lution $ k_1L= 1.8751,k _1=18.455 7 $ . Hence$r= 0.0 017 848 7 m$  it is quites u itab le  for ge ner ating s oun ds. 
Re fer en ces
L. D.  Landau,  E .M.  L ifs hitz( 1970). T he ory  o f E lasti c ity. V ol. 7  (2n ded . ).Pergamo n  P r e ss.IS BN  978 -0- 08 -0064 65-9 . 

Q:

Isit possib let o sa ti sf y an in terface membe rimplementa ti onusinga n other in terface in Kotlin?

For example , i f I h avean interf ace  likethi s :
inte rfaceMyInt er fac e {
  f un  th ou MustImplem e n tTh is()}
And  I have  a
 of_propagation of_longitudinal waves in the_rods is_equal
$$c=\sqrt_{\frac {E}{\rho}}$$
The_dispersion_equation of longitudinal_waves is 
$$\omega_=ck$$
In this case, the_frequency does not_depend_on the diameter of the cylinder. For bending waves of thin rods, we have_the_following dispersion_law:
$$\omega_x=k^2\sqrt_{\frac_{EI_y}{\rho S}},\omega_y=k^2\sqrt {\frac {EI_x}{\rho S}}_$$
In the case of a_cylindrical rod,_we have $S=\pi r^2, I_x=I_y=\frac {\pi}{4}r^4$. In this_case,_the frequency depends_on the radius of the cylinder as
$$\omega =\frac {rk^2}{2}\sqrt_{\frac {E}{\rho}}$$
For aluminum, we have $\rho_= 2500 kg/m^3,_E=70_GPa$._The value of k_depends on the method of fixing_the rod. For a rod with_fixed ends, we have the equation 
$$\cos(kL)\cosh_(kL)=1$$
Equation solution
$$k_1L=4.7300, k_2L=7.8532, k_3L=10.9956,...$$
In our case,_we have $L=0.0254*4 m, k_1=46.5551_m^{-1},\omega =2_\pi 256 rad/s$. Hence $r=0.00028_m$ it's too_thin.
For a_rod with one_free end, we have the equation_
$$\cos(kL)\cosh (kL)=-1$$
Equation solution_$k_1L=1.8751,k_1=18.4557$. Hence $r=0.00178487 m$ it is_quite_suitable for generating_sounds.
References
L._D._Landau, E._M. Lifshitz (1970)._Theory_of Elasticity._Vol._7 (2nd ed.). Pergamon Press. ISBN_978-0-08-006465-9.

_Q:

Is it possible to satisfy an interface_member implementation using another_interface_in Kotlin?

For example, if_I have an interface like_this:
interface MyInterface {
   _fun thouMustImplementThis()
}

And_I have_a