I have a physical pendulum that, for small oscillations, can be modeled with the simple harmonic motion approach. In determining the motion equation, I need to figure out the amplitude: I know that the pendulum has length $L$ and is released from an initial angle of $\theta = 15.0°$. My intuition suggest me that the amplitude can be calculated as $A = L\,\sin\,\theta$, but the solutions book actually uses the formula $A = r\,\theta$ (where in this case $r = L$). But the last formula is used to calculate the arc of circumference, so I wonder if it is the right way to do it and why.
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1$\begingroup$ $sin\theta = \theta$ for small values of $\theta$ but I would not consider $\frac{\pi}{12}$ small enough for that approximation. $\endgroup$– M BarbosaCommented Jun 12, 2016 at 13:12
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4 Answers
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Why would you consider the amplitude to be equal to the horizontal displacement? There's also a vertical displacement, which is arguably more important, without vertical displacement there would be no oscillation, since the potential energy would remain the same. Or take for example a torsion pendulum:
You wouldn't describe the amplitude as $r\sin{\theta}$, because that would give an amplitude of zero when the pendulum swings 360 degrees. It makes more sense to say that the amplitude is equal to the length of the path the pendulum travels, i.e. the length of the arc.
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$\begingroup$ You make some good points about the horizontal and vertical displacements. $\endgroup$ Commented Jun 12, 2016 at 14:03
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Your intuition is correct, $A=L\sin \theta$ is strictly the right answer.
But, note that $\theta$ is (sort of) small. And for small angles: $$\sin \theta \approx \theta$$ The arc length is almost equal to the horizontal distance for small angles.
You even wrote "for small oscillations" in the text yourself, and exactly that is only necessary to add to the discription in order to make this simplification. Then many future formulae that may be derived, are much simpler - but are only valid at small angles.
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$\begingroup$ It's useful to point out that $\theta$ is measured in radians for the small angle formula to work. Beginning students often forget about the proper units. $\endgroup$– Bill NCommented Jun 12, 2016 at 13:33
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1$\begingroup$ @BillN Good point! Of course, SI units are always the case. $\endgroup$– SteevenCommented Jun 12, 2016 at 13:34
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The pendulum would be a simple harmonic oscillator only if the angle from which it is released is very small. It will be actually an angular simple harmonic oscillator but since the angle is small the motion can be considered almost linear. So it can be said a linear simple harmonic oscillator. For large angles, it would be an angular oscillator but not a simple harmonic oscillator. So the only time it makes sense to talk of an amplitude of its harmonic linear motion is when the angles are small. For small angles, $L\sin\theta \approx L\theta $.
If the angle is not small then the oscillation can not be considered as a linear oscillation at all and the only amplitude that would make sense is its angular amplitude - i.e. $\theta$ itself. Or you could consider it oscillating in a plane with the horizontal amplitude being something and vertical amplitude being something else and so on... but that is an unnecessary complication.
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It depends what is being defined as the 'displacement' of the pendulum, which is what you are really asking about. Amplitude is maximum displacement.
The angular displacement $\theta$ of the string or rod could be used. Since the length $L$ is constant (and the CM is assumed to be at the end) this is equivalent to the arc length $s=L\theta$ along the path of the bob.
Alternatively you can define 'displacement' as the horizontal distance $x=L\sin\theta$ of the bob from the vertical line through the pivot.
The confusion is caused (I think) by the fact that the motion is not strictly SHM whichever of $\theta(t)$ or $s(t)$ or $x(t)$ you use. Each is approximately SHM only for 'small' displacements. If one of these were exactly SHM, we would ignore the other options and there would be no confusion.
answered Jun 12, 2016 at 13:38