. Biologists think that some spiders “tune” strands of their web to give enhanced response at
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?
To find the tension in the spider silk strand, we can use the formula for the fundamental frequency of a vibrating string:
f = (1/2L) * sqrt(T/μ)
Where:
f = fundamental frequency (100 Hz)
L = length of the string (12 cm = 0.12 m)
T = tension in the string (unknown)
μ = linear mass density of the string (density * cross-sectional area)
The cross-sectional area of the silk strand can be calculated using the formula for the area of a circle:
A = πr^2
Where:
A = cross-sectional area
r = radius of the silk strand
Given that the diameter of the silk strand is 20 mm, the radius would be 10 mm or 0.01 m.
Substituting the values into the equation for linear mass density, we get:
μ = density * A
μ = 1300 kg/m^3 * (π * (0.01 m)^2)
μ = 0.0407 kg/m
Now, we can substitute all the values into the formula for the fundamental frequency:
100 Hz = (1/2 * 0.12 m) * sqrt(T / 0.0407 kg/m)
Solving for T:
sqrt(T / 0.0407 kg/m) = 100 Hz / (1/2 * 0.12 m)
sqrt(T / 0.0407 kg/m) = 100 Hz / 0.24 m
sqrt(T / 0.0407 kg/m) = 416.67 Hz/m
T / 0.0407 kg/m = (416.67 Hz/m)^2
T / 0.0407 kg/m = 173611.11 Hz^2/m^2
T = 173611.11 Hz^2/m^2 * 0.0407 kg/m
T ≈ 7072.85 N
Therefore, the spider must adjust the tension in the silk strand to approximately 7072.85 Newtons in order to have a fundamental frequency of 100 Hz.