91Ó°ÊÓ

Forced Oscillations and Resonance:

If an external driving force with angular frequency ${\mathrm{Ó\neg }}_d=2\mathrm{Ï€}{f}_d$acts on an oscillating system with natural angular frequency$\mathrm{Ó\neg }=\sqrt{}k\backslash m$the system oscillates with angular frequency ${\mathrm{Ó\neg }}_d.$

The amplitude of the system is largest when

$\mathrm{Ó\neg }d=\mathrm{Ó\neg }$or $2\mathrm{Ï€}{f}_d=\sqrt{\frac{k}{m}}$$(*)$

• The mass of the spider is$m=(2.0\mathrm{g})\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.002\mathrm{kg}$
• The spider has the largest amplitude on its cord if it is tapped once every second.
• We are asked to determine the spring constant of the silk thread.

The spider's amplitude is greatest when the person's driving angular frequency matches the natural angular frequency of the spider's system and the silk thread from the equation.$(*)$

$2\mathrm{Ï€}{f}_d=\sqrt{\frac{k}{m}}$

where$fd$ is the frequency at which the person taps the spider which is equal to$1hz$ since the spider is tapped once each second, $k$is the spring constant of the silk thread, and $m$is the mass of the spider. Solve for $k$

$k=4{\mathrm{Ï€}}^2{f}_d^{2}m$

Subtitle numerical value.

$k=4{\mathrm{Ï€}}^2(1\mathrm{Hz}{)}^2(0.002\mathrm{kg})$

$=0.079\mathrm{N}/\mathrm{m}$