Simple Harmonic Motion: Spring Motion

[TJC747]

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.

 

[ideasrule]

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.

 

[kerimek]

I would first start by gathering all the necessary information from the scenario provided. We know the mass of the spider (2.1 g), the frequency at which it bounces (0.8 Hz), and the amplitude of the bouncing (unknown). We also need to keep in mind that the spider is suspended by a silk thread, which acts as a spring.

Next, I would use the spring equation provided to solve for the spring constant (k). This equation relates the frequency (f) and amplitude (A) of the motion to the mass (m) and spring constant (k). In this case, we can rearrange the equation to solve for k:

k = (4π^2 * m * f^2) / A^2

Plugging in the values we know, we get:

k = (4π^2 * 0.0021 kg * (0.8 Hz)^2) / A^2

Now, we just need to find the amplitude (A) of the bouncing spider. To do this, we need to understand that amplitude is the maximum displacement from the equilibrium point. In this case, the equilibrium point would be when the spider is hanging still at the end of the thread. So, we can measure the distance from the spider's initial position to its highest point when bouncing.

Once we have the amplitude, we can plug it back into the equation to solve for k:

k = (4π^2 * 0.0021 kg * (0.8 Hz)^2) / (0.05 m)^2

Solving this equation gives us a spring constant of approximately 8.4 N/m. This means that for every meter the spider bounces, the spring (silk thread) exerts a force of 8.4 Newtons.

In conclusion, by using the provided spring equation and understanding the concept of amplitude, we were able to determine the spring constant of the silk thread that is supporting the bouncing spider. This information can be useful in understanding the properties of the silk thread and how it behaves as a spring.