Biologists think that some spiders “tune” strands of their web to give enhanced response at

Question

frequencies corresponding to those at which desirable prey might struggle. Orb spider web
silk has a typical diameter of 20 mm 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?

Answer 1

To calculate the tension in the spider's silk strand, we can use the equation for the fundamental frequency of a string:

f = (1/2L) * sqrt(T/μ)

Where:
f = fundamental frequency
L = length of the string
T = tension in the string
μ = linear mass density

Given:
Length of the silk strand (L) = 12 cm = 0.12 m
Diameter of the silk strand = 20 mm = 0.02 m
Radius of the silk strand (r) = (0.02 m) / 2 = 0.01 m
Density of spider silk (μ) = 1300 kg/m^3
Fundamental frequency (f) = 100 Hz

We can calculate the linear mass density using the formula:

μ = (m/L) * πr^2

Where:
m = mass of the silk strand

To calculate the mass (m), we can use the formula for the volume of a cylinder:

V = πr^2h

Given:
Length of the silk strand (h) = 0.12 m
Radius of the silk strand (r) = 0.01 m

V = π(0.01 m)^2 * 0.12 m
V ≈ 3.769 x 10^-6 m^3

Since density (μ) = mass/volume, we can rearrange the equation to solve for mass (m):

m = μ * V
m = 1300 kg/m^3 * 3.769 x 10^-6 m^3
m = 4.8937 x 10^-3 kg

Then, we can calculate the linear mass density:

μ = (m/L) * πr^2
μ = (4.8937 x 10^-3 kg / 0.12 m) * π(0.01 m)^2
μ ≈ 0.4267
μ ≈ 0.43 kg/m

Now, we can rearrange the equation for the fundamental frequency to solve for tension (T):

f = (1/2L) * sqrt(T/μ)
T = (f * 2L)^2 * μ
T = (100 Hz * 2 * 0.12 m)^2 * 0.43 kg/m
T ≈ 316.23 N

Therefore, the spider must adjust the tension in the silk strand to approximately 316.23 N in order to have a fundamental frequency of 100 Hz.