Question

Biologists think that some spiders “tune” strands of their web to give an enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20 µm, and spider silk has a density of 1300 kg/m³. To have a fundamental frequency at 100 Hz, to what tension must a spider adjust a 12-cm-long strand of silk?

Short answer

The tension in the silk strand is 2.35 x 10^-4 N

Step-by-step solution

Write the given information

Diameter of spider web d= 20 µm =20 x 10^-6 m
Radius of the spider web, r= 10^-5m
The density of the spider silk, ρ= 1300 kg/ m³
The frequency of the silk strand, f = 100Hz
Length of the silk strand, L= 12 cm= 0.12m

 Step 2: To determine the tension in the silk

The mass per unit length of the silk is given by
$$\begin{gathered} \mu =\rho A \\ \mu =\rho \left(\pi r^2\right) \\ =1300\left\{3.14{\left({10}^{-5}\right)}^2\right\} \\ =4.082x{10}^{-7}kg/m \end{gathered}$$

The fundamental frequency of the wave is given by
localid="1650099139558" $f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

Substitute the known values

localid="1650099446065" $$\begin{gathered} f=\frac{1}{2L}\sqrt{\frac{T}{\mu }} \\ 100=\frac{1}{2(0.12)}\sqrt{\frac{T}{4.082x{10}^{-7}kg/m}} \\ 24=\sqrt{\frac{T}{4.082x{10}^{-7}}} \end{gathered}$$

Squaring on both the sides
$$\begin{gathered} 576=\frac{T}{4.082x{10}^{-7}} \\ T=2.35x{10}^{-4}N \end{gathered}$$

Thus the tension in the silk strand is 2.35 x 10^-4 N