Spiders may "tune" strands of their webs to give enhanced response at frequencies corresponding to the frequencies at which desirable prey might struggle. Orb web silk has a typical diameter of 0.0020 mm and spider silk has a density of 1300kg/m^3. To give a resonance at 200Hz to what tension must a spider adjust a 16cm long strand of silk?

Respuesta :

Explanation:

Formula for wave velocity is as follows.

           [tex]\nu = \sqrt{\frac{T}{\frac{m}{L}}}[/tex] .......... (1)

Now, we know that relation between velocity and frequency is as follows.

           f = [tex]\frac{\nu}{2L}[/tex] ......... (2)

Now, on equating equation (2) in equation (1) is as follows.

             f = [tex]\frac{\sqrt{\frac{T}{\frac{m}{L}}}}{2L}[/tex]

We will calculate the mass with the help of relation between density and volume as follows.

            Density [tex](\rho) = \frac{m}{V}[/tex]

                 m = [tex]\rho (AL)[/tex]

                     = [tex]1300 kg/m^{3} \times \frac{0.002 \times 10^{-3}}{2} \times 3.14 \times 0.14[/tex]

                     = [tex]5.71 \times 10^{-10} kg[/tex]

Now, we will rewrite the equation f = [tex]\frac{\sqrt{\frac{T}{\frac{m}{L}}}}{2L}[/tex] for T as follows.

             [tex](2Lf)^{2} = \frac{T}{\frac{m}{L}}[/tex]

                      T = [tex]\frac{m}{L} (2Lf)^{2}[/tex]

                          = [tex][\frac{5.71 \times 10^{-10} kg}{0.16 m}](2 (0.16 m)(100 Hz))^{2}[/tex]

                         = [tex]3.65 \times 10^{-6} N[/tex]

thus, we can conclude that up to [tex]3.65 \times 10^{-6} N[/tex] tension must a spider adjust a 16cm long strand of silk.