Question

$I$A $2.0g$ spider is dangling at the end of a silk thread. You can make the spider bounce up and down on the thread by tapping lightly on his feet with a pencil. You soon discover that you can give the spider the largest amplitude on his little bungee cord if you tap exactly once every second. What is the spring constant of the silk thread?

Short answer

The spring constant of the silk thread is$k=0.079\mathrm{N}/\mathrm{m}$

Step-by-step solution

Concept and principle.

Forced Oscillations and Resonance:

If an external driving force with angular frequency ${\omega }_d=2\pi f_d$acts on an oscillating system with natural angular frequency$\omega =\sqrt{}k\backslash m$the system oscillates with angular frequency ${\omega }_d.$

The amplitude of the system is largest when

$\omega d=\omega$or $2\pi f_d=\sqrt{\frac{k}{m}}$$(*)$

Given data.

• 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.

Solution 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.$(*)$

$2\pi f_d=\sqrt{\frac{k}{m}}$

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 $k$

$k=4{\pi }^2{f}_d^{2}m$

Subtitle numerical value.

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

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