Question

2. 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 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?

Answer 1

Answer:

The tension is 2.35x10⁻⁴N

Explanation:

Given:

d = 20 μm = 20x10⁻⁶ m

L = 12 cm = 0.12 m

f = 100 Hz

The transverse velocity is:

$v=\sqrt{\frac{T_{s} }{\mu } }$ (eq. 1)

Where

Ts = web´s tension

μ = linear mass density

The linear mass density is:

$\mu =\frac{\pi d^{2}\rho }{4}$ (eq. 2)

Where

ρ = density of the spider silk = 1300 kg/m³

The fundamental frequency is:

$f=\frac{v}{2L}$

Replacing eq. 1 and 2 and clearing the tension Ts:

$f=\frac{1}{2L} \sqrt{\frac{4T_{s} }{\pi d^{2}\rho } } \\T_{s} =\pi d^{2} f^{2} L^{2} \rho$

$T_{s} =\pi *(20x10^{-6} )^{2}*(100 )^{2} *(0.12 )^{2}*1300=2.35x10^{-4} N$

Answer 2

Answer: 2.35*10^-4 N

Explanation:

The linear mass density is

μ = m / L

Recall, ρ = m / V, so that, m = ρV

Substitute this in the equation

μ = ρV / L

Also, recall, V = AL. Substituting this in the equation, we have

μ = ρAL / L = ρA

Again, recall, A = πr², so that

μ = ρπr²

μ = 1300 * 3.142 * [10*10^-6]²

μ = 4.085*10^-7 kg/m

the wavelength of the wave in the silk, λ = 2L

λ = 2 * 0.12 = 0.24 m

The wave speed, v = fλ

v = 100 * 0.24 = 24 m/s

v = √(T / μ)

v² = T / μ

T = v²μ

T = 24² * 4.085*10^-7

T = 2.35*10^-4 N