Simple harmonic motion of a pendulum

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Homework Statement



I need to find the phase difference between two pendulums. My problem is that I don;t seem to have enough information to calculate the phase for the second pendulum (I have already calculated the phase for the first).

The pendulum is 0.35m long, and is at 0.00rad at t=0. (the question also states that the pendulum IS swinging). I need to find the phase at 10s


Homework Equations



θ(t)=[itex]\Theta[/itex]sin(ωt+[itex]\phi[/itex])

θ(t)=[itex]\Theta[/itex]sin([itex]\phi[/itex])

phase = ωt+[itex]\phi[/itex]

ω=[itex]\sqrt{g/l}[/itex]

[itex]\phi[/itex]=arcsin(θ(t)/[itex]\Theta[/itex])



The Attempt at a Solution



So as I mentioned, I have calculated the phase of the first pendulum. But I cant seem to calculate the phase of the second one.

I thought there might have been a way to calculate [itex]\Theta[/itex] and then calculate [itex]\phi[/itex] using that value but can't seem to figure out exactly how!

HELP!! :)
 

Answers and Replies

  • #2
cepheid
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Homework Statement



I need to find the phase difference between two pendulums. My problem is that I don;t seem to have enough information to calculate the phase for the second pendulum (I have already calculated the phase for the first).

The pendulum is 0.35m long, and is at 0.00rad at t=0. (the question also states that the pendulum IS swinging). I need to find the phase at 10s


Homework Equations



θ(t)=[itex]\Theta[/itex]sin(ωt+[itex]\phi[/itex])

θ(t)=[itex]\Theta[/itex]sin([itex]\phi[/itex])

phase = ωt+[itex]\phi[/itex]

ω=[itex]\sqrt{g/l}[/itex]

[itex]\phi[/itex]=arcsin(θ(t)/[itex]\Theta[/itex])



The Attempt at a Solution



So as I mentioned, I have calculated the phase of the first pendulum. But I cant seem to calculate the phase of the second one.

I thought there might have been a way to calculate [itex]\Theta[/itex] and then calculate [itex]\phi[/itex] using that value but can't seem to figure out exactly how!

HELP!! :)
Welcome to PF,

You know that [itex]\phi[/itex] is either 0 or π, because this is given (implicitly) in the problem. If, at t = 0, the angular position of the pendulum is 0, then you have:

[tex] \theta(0) = \Theta \sin(0 + \phi) = 0 [/tex]

[tex] \Rightarrow \sin(\phi) = 0 [/tex]

[tex] \Rightarrow \phi = 0~~\text{or}~~\pi [/tex]

assuming the initial phase is specified to be an angle somewhere in the interval [0, 2π).

So we have either:

[tex] \theta(t) = \Theta \sin(\omega t ) [/tex]
[tex] \theta(t) = \Theta \sin(\omega t + \pi ) = -\Theta \sin(\omega t ) [/tex]

It should be pretty intuitive to you that the motion in this case would be described by a pure sine function (with no phase offset). Think about what sin(x) does at x = 0: it is crossing the zero point (the point midway between the maximum and minimum values of the oscillation), and it is doing so with a derivative that is maximally positive. In other words, if the motion is described by a sine function, then at t = 0 the pendulum is crossing the midpoint with maximum positive speed. Now look at a plot of -sin(x). It's the same, except that the slope is maximally negative at x = 0 instead of maximally positive. So if the motion is described by the negative of a sine function, then at t = 0 the pendulum is crossing the midpoint with maximum negative speed. So, to choose between the two, you need another initial condition, namely the direction of the initial velocity (you already know that its magnitude must be equal to the max speed).

I suspect that you have the freedom to choose which direction is positive and which is negative, in which case you could pick either +sin(ωt) or -sin(ωt).
 

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