Question

"

A small Insect 0.25g insert a caught in a spider web.The web vibrated at 20 Hz. 1) What is the spring constant of the web? 2) What frequency will the web vibrate if an insect with a mass of 0.10g?

Answer 1

Approximately
`3.95\; {\rm N\cdot m^(-1)}`.

Approximately
`31.6\; {\rm Hz}` when mass is
`0.10\; {\rm g}`.

Step-by-step explanation:

Assume that the mass of the web is negligible, and that the insect is vibrating horizontally. Model the system of the vibrating insect and the web as a mass on a frictionless surface attached to a horizontal spring. Analyze the equations for a mass-spring system in simple harmonic motion to find the relationship between the spring constant, mass, and frequency.

In a frictionless horizontal mass-spring system, the net force on the mass is equal to the restoring force from the spring. When the mass is at a position of
`x` relative to the equilibrium position, the net force on the mass would be:

`\begin{aligned}& (\text{net force}) \\ =\; & (\text{restoring force}) \\ =\; & -k\, x \end{aligned}`.

(Negative because restoring force is opposite in direction to displacement.)

Acceleration of the mass
`m` would be:

`\begin{aligned}a &= \frac{(\text{net force})}{m} = (-k\, x)/(m)\end{aligned}`.

Let
`k` denote the spring constant, let
`f` denote the frequency of the motion, and let
`\omega = 2\, \pi\, f` denote the angular velocity. In a simple harmonic motion, the acceleration
`a` and displacement
`x` at time
`t` can be modelled as:

• 
  `x(t) = A\, \cos(\omega t)`, and
• 
  `a(t) = -A\, \omega^(2)\, \cos(\omega\, t)`.

Where
`A` is the amplitude of the motion (
`A > 0`.)

Since
`a = (-k\, x ) / m`, substitute in the expression for
`x(t)` and
`a(t)` to obtain:

`\displaystyle -A\, \omega^(2)\, \cos(\omega\, t) = (-k\, A\, \cos(\omega\, t))/(m)`.

`\displaystyle -A\, (2\, \pi\, f)^(2)\, \cos(2\, \pi\, f\, t) = (-k\, A\, \cos(2\, \pi\, f\, t))/(m)`.

Simplify this expression (assuming that
`\cos(2\, \pi\, f\, t) \\e 0`) to obtain:

`\displaystyle (2\, \pi\, f)^(2) = (k)/(m)`.

Solve for the spring constant
`k`:

`\begin{aligned}k &= (2\, \pi\, f)^(2)\, m\end{aligned}`.

Apply unit conversion and ensure that mass is measured in standard units (kilograms):

`\begin{aligned} m &= 0.25\; {\rm g} * \frac{1\; {\rm kg}}{1000\; {\rm g}}\\ &= 2.5 * 10^(-4)\; {\rm kg}\end{aligned}`.

Given that
`f = 20\; {\rm Hz}`:

`\begin{aligned}k &= (2\, \pi\, f)^(2)\, m \\ &= (2\, \pi\, (20\; {\rm Hz}))^(2)\, (2.5 * 10^(-4)\; {\rm kg}) \\ &\approx 3.95 \; {\rm kg \cdot s^(-2)} \\ &= 3.95\; {\rm N\cdot m^(-1)}\end{aligned}`.

In other words, the equivalent spring constant of the web is approximately
`3.95 \; {\rm N\cdot m^(-1)}`.

In the other part of the question, the goal is to find frequency given mass
`m = 0.10\; {\rm g} = 1.0* 10^(-4)\; {\rm kg}` (and that
`k \approx 3.95 \; {\rm N\cdot m^(-1)}`.) Rearrange the previous equation to find an expression for frequency
`f` in terms of mass
`m` and spring constant
`k`:

`\begin{aligned}k &= (2\, \pi\, f)^(2)\, m\end{aligned}`.

`\displaystyle k = (2\, \pi)^(2)\, f^(2)\, m`.

`\displaystyle f^(2) = (k)/((2\, \pi)^(2)\, m)`.

`\begin{aligned} f &= \sqrt{(k)/((2\, \pi)^(2)\, m)} \\ &\approx \sqrt{\frac{3.9478\; {\rm N\cdot m^(-1)}}{(2\, \pi)^(2)\, (1.0* 10^(-4)\; {\rm kg})}} \\ &\approx 31.6\; {\rm s^(-1)} \\ &\approx 31.6\; {\rm Hz}\end{aligned}`.