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I've reading about Simple harmonic motion, and it is said that SHM occurs only when there is force $F \propto -kx$. This is true for spring-mass systems. However, in systems like the Pendulum, we have the torque $\tau$ which is proportional to angular displacement, such that $\tau \propto -\kappa\theta$.

So, even though it is the angular displacement $\theta$, instead of $x$, that is undergoing sinusoidal motion, is it okay to call this motion 'Simple Harmonic' ? If so, then shouldn't it be wrong to say that SHM occurs only when 'Force' is inversely proportional to 'Distance'

For example, in Series and Parallel LC circuits, you have oscillations of charge in the capacitor, that has a sinusoidal form exactly analogous to the displacement $x$ in a spring mass system. However, we don't call this SHM.

When exactly is the terminology $SHM$ applicable when describing an oscillatory system ? If one says, that it is only applicable, if the oscillating variable is the displacement, then shouldn't pendulum or torsional pendulum be disqualified from being labeled as $SHM$, since it is the angular displacement that oscillates in their case, instead of the ordinary displacement $x$?

However, I'm also led to believe that torque is generated by force, so in some sense force is more fundamental than torque. In a torsional pendulum, does $\tau\propto-\kappa\theta$ automatically imply some $F\propto -kx$ ? If so, where is this force coming from, and can we analyze a torsional pendulum, purely using forces without using any kind of torque?

Is SHM just one of the analogous systems that can be modeled by the more general Harmonic oscillator terminology?

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  • $\begingroup$ @RayPalmer When you say "Pendulum" are you talking about a mass pivoted about a point and swinging freely in a gravitational field? Or a torsional pendulum, ignoring gravity? $\endgroup$
    – Bill N
    Commented Nov 1, 2021 at 12:30
  • $\begingroup$ You can describe the motion of the pendulum in terms of force and linear displacement if you want. It looks like your question is more about labeling. The physics is the same no matter how you label it. $\endgroup$
    – nasu
    Commented Nov 1, 2021 at 16:27
  • $\begingroup$ There are a great many cases when the mathematics of simple harmonic motion comes into play if only as an approximation. Almost any place where you have a zero first derivative and negative second derivative, you effectively have a restorative "force" even if the quantities in question have little to do with motion. $\endgroup$
    – R. Romero
    Commented Nov 1, 2021 at 16:50

3 Answers 3

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The mathematical representation of the particle in simple harmonic motion model is based on three separate equations, not only one (which you have pointed out): $$F = -kx$$ $$\frac{d^2x}{dt^2} = -\omega^2x$$ $$x(t) = A\text{cos}(\omega t+\phi)$$

Hence if you are analyzing a mechanical situation and you find that it is of the form of any of these equations, it's motion can be modelled as simple harmonic.

In the case of the torsional pendulum, it's position can be described as: $$\tau = I\frac{d^2\theta}{dt^2} = -\kappa\theta$$ which is analogous to the second equation and the situation is a mechanical one. Hence, as long as the elastic limits of the wire isn't exceeded, this motion can be modelled as simple harmonic.

An LC circuit cannot be strictly modelled as simple harmonic, because it is not a mechanical one. However, the oscillations of the value of electric and magnetic field in the capacitor and inductor very closely resemble a spring-block system. This fact is used to obtain the equation: $$q = Q_{\text{max}} \text{cos}(\omega t + \phi)$$

Also note

'Force' is inversely proportional to 'Distance'

The term $x$ here refers to position, relative to the equilibrium position, and not strictly the distance.

Hope this helps.

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  • $\begingroup$ So, strictly speaking, saying that SHM is possible only when Force is proportional position is somewhat wrong. In case of the pendulum, we have the Torque proportional to the angular position instead. Unless torque proportional to angular position automatically implies that force is proportional to position - the way SHM is usually defined in the books, is strictly speaking, misleading $\endgroup$
    – RayPalmer
    Commented Oct 31, 2021 at 16:34
  • $\begingroup$ @RayPalmer, it's not misleading because a particle that has a force $-kx$ exerted on it does exhibit simple harmonic motion. It's just that there are also two more equations which if satisfied ( or analogous to ), can form the basis for the particle being in simple harmonic motion. In the case of torque, we have angular acceleration that is proportional to angular position. $\endgroup$
    – Cross
    Commented Oct 31, 2021 at 16:45
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    $\begingroup$ It doesn't have to be motion to be simple harmonic. The LC circuit does have simple harmonic oscillations because the DE matches the pattern $y''=-\beta y$. It's the DE which defines simple harmonic behavior. Your other two equations are redundant as far as a definition is concerned, i.e., they are not independent equations. The force equation is a physics expression of the DE, the $x(t)=$ equation is the solution to the DE. $\endgroup$
    – Bill N
    Commented Nov 1, 2021 at 12:37
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I think that most physicists use simple harmonic to describe a variation in some quantity, $\xi$, (be it linear displacement, angular displacement or even electric current) that is sinusoidal with respect to time. In other words $$\xi=\xi_0 \sin(\omega t - \phi)$$ in which $\xi_0$, $\omega$ and $\phi$ are constants.

Alternatively we could say that $\xi$ has to be a real quantity obeying the equation $$\frac{d^2 \xi}{dt^2}=-\omega^2 \xi$$ A periodic but non-sinusoidal variation would, on account of Fourier's theorem, be a "compound harmonic motion", though the term is seldom used.

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  • $\begingroup$ So, $\xi$ can represent any physical quantity, and if it satisfies the differential equation, we would call it simple harmonic ? Hook's law is just a special case, where if you write it in differential form, the displacement $x$, happens to fit the differential equation. So, we usually say hook's law is needed for simple harmonic motion. In truth, any variable that satisfies that equation undergoes SHM. More like, if you want displacement to undergo SHM, you need hook's law. $\endgroup$
    – RayPalmer
    Commented Oct 31, 2021 at 18:35
  • $\begingroup$ In case of the torsional pendulum, it is the angular displacement $\theta$, that satisfies the differential equation, as $\tau\propto-\kappa\theta$, and this is considered to be the rotational analog of Hook's law. My question is, is this derived from Hook's law, and just an easier way to express the Hook's law in this scenario ? Or is it an independent observation, and an independent law analogous to hook's law ? $\endgroup$
    – RayPalmer
    Commented Oct 31, 2021 at 18:38
  • $\begingroup$ Please tell me what the "this" at the end of your second line, and the "it" near the end of your third line refer to. $\endgroup$ Commented Oct 31, 2021 at 19:13
  • $\begingroup$ 'This' at the end of second line and 'it' at the end of the third line, both refer to the equation $\tau\propto-\kappa\theta$. I know that torque is created by force, but the torsional pendulum is always analyzed using torques, and spring mass system is analyzed using forces. If Hook's law represents restoring force, and the equation above represents restoring torque, and we know that torque is created by force, can we derive the above equation from the force equation of hook's law ? $\endgroup$
    – RayPalmer
    Commented Oct 31, 2021 at 19:26
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    $\begingroup$ Can the proportionality relation between torque and angle of twist (let's say for a wire) be derived from a proportionality relation between force and extension? Not quite. We can derive the torque/angle relation by considering forces, but these are tangential or shear forces, giving rise to a 'side slip' or shear strain. In other words we need a generalised Hooke's law rather than one involving longitudinal forces and extensions. $\endgroup$ Commented Oct 31, 2021 at 19:55
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In the case of the simple pendulum, this is usually referred to as approximately simple harmonic motion, and is only regarded as valid for small angles of deflection. This is because the equation of motion is $$l\ddot{\theta}=-g\sin\theta$$ and when $\theta$ is sufficiently small, we can make the approximation $$\sin\theta\simeq\theta,$$thus giving the SHM equation in its standard form.

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