The surface of the web woven by the spider is horizontal. When the spider is in its center, the web deflects, the value of deflection $x_{0}=1 mm$. A fly, once caught in the web, starts beating. Can we determine from the frequency of resonance oscillations whether the mass of the fly exceeds the mass of the spider or not?
Decision:
The web is an elastic system with zero mass. Considering that for small stretches of the web, the forces acting in it obey Hooke's law, we can write the following equation:
$m_{0}g=kx_{0}$. (1)
Here $m_{0}$ is the mass of the spider;g is the acceleration of free fall; k is the coefficient of elasticity of the web. A web loaded by a spider of mass $m_{0}$ and a web loaded by a fly of mass $m$ are different oscillating systems. The circular frequencies $\omega_{0}$ and $\omega$ of the natural (resonant) oscillations of these systems are defined by the formulas
$\omega_{0} = \frac{2 \pi}{T_{0}} = \sqrt{ \frac{k}{m_{0}}},\omega= \frac{2 \pi }{T} = \sqrt{ \frac{k}{m}}$. (2)
Here $T_{0}$ and $T$ are the periods of natural oscillations of the systems. From the system of equations (2) it is not difficult to obtain the equality
$\frac{m}{m_{0}} = \frac{ \omega^{2}_{0}}{ \omega^{2}}$, (3)
which contains an unknown quantity $\omega_{0}$. The latter is not difficult to find it by excluding with the help of (1) from the first equality (2) the relation $\frac{k}{m_{0}}$
$\omega_{0} = \sqrt{ \frac{g}{x_{0}}}$. (4)
Substituting (4) into (3), we find
$\frac{m}{m_{0}} = \frac{g}{(x_{0}\omega^{2}}$.
From this it is clear that if the mass of the fly exceeds the mass of the spider, then
$\omega < \sqrt{ \frac{g}{x_{0}}}$,
but given the numerical data of the problem, it's equivalent to the condition
$\omega < 100 c^{-1}$.
Thus, the frequency of resonance oscillations of the web with the fly can be used to determine whether the mass of the fly exceeds the mass of the spider or not.
Decision:
The web is an elastic system with zero mass. Considering that for small stretches of the web, the forces acting in it obey Hooke's law, we can write the following equation:
$m_{0}g=kx_{0}$. (1)
Here $m_{0}$ is the mass of the spider;g is the acceleration of free fall; k is the coefficient of elasticity of the web. A web loaded by a spider of mass $m_{0}$ and a web loaded by a fly of mass $m$ are different oscillating systems. The circular frequencies $\omega_{0}$ and $\omega$ of the natural (resonant) oscillations of these systems are defined by the formulas
$\omega_{0} = \frac{2 \pi}{T_{0}} = \sqrt{ \frac{k}{m_{0}}},\omega= \frac{2 \pi }{T} = \sqrt{ \frac{k}{m}}$. (2)
Here $T_{0}$ and $T$ are the periods of natural oscillations of the systems. From the system of equations (2) it is not difficult to obtain the equality
$\frac{m}{m_{0}} = \frac{ \omega^{2}_{0}}{ \omega^{2}}$, (3)
which contains an unknown quantity $\omega_{0}$. The latter is not difficult to find it by excluding with the help of (1) from the first equality (2) the relation $\frac{k}{m_{0}}$
$\omega_{0} = \sqrt{ \frac{g}{x_{0}}}$. (4)
Substituting (4) into (3), we find
$\frac{m}{m_{0}} = \frac{g}{(x_{0}\omega^{2}}$.
From this it is clear that if the mass of the fly exceeds the mass of the spider, then
$\omega < \sqrt{ \frac{g}{x_{0}}}$,
but given the numerical data of the problem, it's equivalent to the condition
$\omega < 100 c^{-1}$.
Thus, the frequency of resonance oscillations of the web with the fly can be used to determine whether the mass of the fly exceeds the mass of the spider or not.