Simple Harmonic Motion: Spring Motion

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A 2.1 g 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 at a frequency of 0.8 Hz. What is the spring constant of the silk thread?
(answer to be in N/m)

Certainly understanding the spring equation that states v = sqrt(k/m)*sqrt(A^2 - X^2)
But how to apply it? I'm clueless. Help would be appreciated. Thanks.
 

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  • #2
ideasrule
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8 Hz is the thread's resonant frequency: the spider, if let alone at a point above the equilibrium point, would naturally oscillate at 8 Hz. This is because 8Hz taps would hit the spider at the same point in its swing every time, so the taps reinforce each other instead of cancelling.
 

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