Question

Spiders may "tune" strands of their webs to give enhanced response at frequencies corresponding to the frequencies at which desirable prey might struggle. Orb web silk has a typical diameter of 0.0020 mm and spider silk has a density of 1300kg/m^3. To give a resonance at 200Hz to what tension must a spider adjust a 16cm long strand of silk?

Answer 1

Step-by-step explanation:

Formula for wave velocity is as follows.

`\\u = \sqrt{(T)/((m)/(L))}` .......... (1)

Now, we know that relation between velocity and frequency is as follows.

f =
`(\\u)/(2L)` ......... (2)

Now, on equating equation (2) in equation (1) is as follows.

f =
`\frac{\sqrt{(T)/((m)/(L))}}{2L}`

We will calculate the mass with the help of relation between density and volume as follows.

Density
`(\rho) = (m)/(V)`

m =
`\rho (AL)`

=
`1300 kg/m^(3) * (0.002 * 10^(-3))/(2) * 3.14 * 0.14`

=
`5.71 * 10^(-10) kg`

Now, we will rewrite the equation f =
`\frac{\sqrt{(T)/((m)/(L))}}{2L}` for T as follows.

`(2Lf)^(2) = (T)/((m)/(L))`

T =
`(m)/(L) (2Lf)^(2)`

=
`[(5.71 * 10^(-10) kg)/(0.16 m)](2 (0.16 m)(100 Hz))^(2)`

=
`3.65 * 10^(-6) N`

thus, we can conclude that up to
`3.65 * 10^(-6) N` tension must a spider adjust a 16cm long strand of silk.