Answer:
T=2.94*10^-10 N/m.
Step-by-step explanation:
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 20μm, and spider silk has a density of 1300 kg/m³.
To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?
l=length of the spider silk, 14cm
velocity of wave = √(T/μ)
where T = tension and
μ = mass per unit length)
λ/2=l
for fundamental frequency λ/2 =14cm
(λ= wavelength of standing wave; as there will be no node
except the endpoints of silk strand)
λ = 28 cm = 0.28 m
and since frequency * wavelength = speed of wave. we have,
150 * 0.28 = √(T/μ) ..................(#)
now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density
= [π * (10 * 10^(-6))²] * 1300 = 13π * 10^(-8).
now putting this in equation (#) we get
150 * 0.28 = √(T/[13π * 10^(-8)]).
thus T = [13π * 10^(-8)] * (42)² =
2.94*10^-10 N/m.