Oh yes, sorry I should have shown the first part too, my bad.
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This is just a brainstorm here:
Assume that the angular displacement goes on forever, i.e. after it reaches \(2\pi\) it'll keep going to \(4\pi\), \(6\pi\) and so on.
Using \(v = r\omega\) you can deduce that \(\omega = \sqrt{ \frac{g\tan \theta}{R} } \). Using \( \theta = \omega t\) you can then deduce that for each particle, \( \theta = t \sqrt{ \frac{g\tan \theta}{R} } \). Note that since the particles start side by side, we just naturally assume that their initial angular displacement is \(\theta = 0\) for convenience.
By subbing in \(g=10\) along with both corresponding values of \(\tan\theta\) and \(R\), you can deduce which of the two particles is travelling faster.
Then, the first time the particles are next to each other should be when \( \boxed{ \theta_{faster} - \theta_{slower} = 2\pi}\). In fact, the second time they are next to each other should also be when this difference equals to 4*pi. Essentially, at each point where their angular displacements
differ by some integer multiple of \(2\pi\), they are next to each other.
Hi I need help with this question below on how to approach it and the logic behind the steps. Thank you
Sketch y=f(x)=(x^n)e^-x for x>0, n>1
Thank you
This question is simply screaming "multiplication of \(y\)-coordinates" and that's all there is to it.
Are you uncomfortable with the whole process of multiplying ordinates and require that logic to be expanded for your example?