Mass hanging by spring tracing a eight shaped curve

AI Thread Summary
A mass on a spring executes up-and-down vibrations while swinging, creating a figure-eight motion. The problem requires finding the spring's force constant in terms of mass, length, and gravitational acceleration. The derived formula for the spring constant is k = 4mg/l. The relationship between the oscillation frequencies of the spring and pendulum is crucial, as the spring's frequency is twice that of the pendulum. Understanding this dynamic is key to solving the problem effectively.
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Homework Statement


A mass ##m## on the end of a light spring of force constant ##k## stretches the spring to a length ##l## when at rest. The mass is now set into motion so it executes up and down vibrations while swinging back and forth as a pendulum. The mass moves in a figure-eight pattern in a vertical plane, as shown in the figure. Find the force constant in terms of ##m##,##l## and ##g##.

(Ans: k=4mg/l )

Homework Equations


The Attempt at a Solution


I noticed that the curve traced by the hanging mass is of the form ##r^2=a\cos(2\theta)## with the mass ##m## being at the origin at ##t=0##. But I don't think this is going to help. This is a question from one of my practice sheets and I doubt I need to deal with polar curves which aren't generally taught in high school.

How do I approach this problem? :confused:

Any help is appreciated. Thanks!
 

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Since it traces out a figure eight that would suggest that one oscillation is twice what the other oscillation is so maybe you can factor that into your solution.
 
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Lets me think of Lissajous figures. For a figure 8 the frequencies of oscillations are 1:2. In this case the up-down oscillations are twice those of the pendulum motion. Not sure this helps or is relevant though.
 
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jedishrfu said:
Since it traces out a figure eight that would suggest that one oscillation is twice what the other oscillation is so maybe you can factor that into your solution.

Ah yes, thanks a lot jedishrfu! :)

$$2\times 2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{l}{g}}$$
$$\Rightarrow k=\frac{4mg}{l}$$
 
I think the springs' frequency of oscillations is twice that of the pendulum in this case.
 
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