I've been taught that in a simple pendulum, for small $x$, $\sin x \approx x$. We then derive the formula for the time period of the pendulum. But I still don't understand the Physics behind it. Also, there's no angle $x$ involved in a spring-mass system, then why do we consider it an SHM only for small amplitudes?

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    $\begingroup$ I've deleted some comments that should have been answers, along with their responses. $\endgroup$
    – David Z
    Commented Feb 22, 2017 at 6:34
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    $\begingroup$ @DavidZ: isn't that a bit hasty; shouldn't they have been prompted to turn their comments into answers? $\endgroup$ Commented Feb 22, 2017 at 13:39
  • $\begingroup$ @MoziburUllah Sometimes we remind people, but not always. In this case the popularity of the question is one factor that pushes toward deleting answer-like comments sooner rather than later. $\endgroup$
    – David Z
    Commented Feb 22, 2017 at 20:29
  • $\begingroup$ @DavidZ: I can't say I follow you reasoning; one reason I put it as a comment rather than as an answer was simply that it was short. $\endgroup$ Commented Feb 22, 2017 at 23:18
  • $\begingroup$ @MoziburUllah That's not a reason to post something as a comment rather than an answer. Answers can be short too. I would be happy to explain more about when to use comments vs answers, if you like, in Physics Chat. $\endgroup$
    – David Z
    Commented Feb 22, 2017 at 23:31

3 Answers 3


A simple pendulum does not strictly show simple harmonic motion unless you allow some approximations and uncertainties. It approximately behaves as a harmonic oscillator for small amplitudes.

An object is said to be executing simple harmonic motion (no damping; not a forced oscillation) if and only if it satisfies the following condition:

$$\frac{d^2 \phi}{dt^2} = -\omega^2 \phi \tag{1}$$

where $\phi$ is a variable quantity such as displacement, angular displacement, etc.

Does a pendulum execute simple harmonic motion?

enter image description here

The equation of motion for the pendulum can be written as:

$$\vec{F} = {m\vec{g}} - \vec{T}$$

We know that the pendulum bob will move in a circle (assume that the string does not stretch), therefore, there is no motion in the direction of the string. This would mean that the net force on the bob will be used to provide a constant centripetal force.

$$F_{radial} = T - mg\cos \theta = \frac{mv^2}{L}$$

The acceleration along the circumference of the string can be written as:

$$F_{tangential} = ma = mg \sin \theta$$

$$a_{tangential} = a = g \sin \theta \tag{2}$$

The tangential acceleration can be expressed in terms of the angle $\theta$ as follows:

$$v = L \frac{d\theta}{dt}$$ $$\frac{dv}{dt} = a = -L\frac{d^2\theta}{dt^2} \tag{3}$$

We have a minus sign because the gravitational force (acceleration) always tries to decrease the angle $\theta$.

Substituting $(3)$ in $(2)$, you get,

$$L\frac{d^2\theta}{dt^2} = -g \sin \theta \tag{4}$$

If you compare equation $(4)$ with equation $(1)$, you'll notice that it does not match.

This would mean that the pendulum bob does not execute a simple harmonic motion.

However, if the amplitude is small, then the maximum value of $\theta$ is small. The small angle approximation can be stated as follows:

$$\sin \theta \approx \theta$$

enter image description here

Image Source: Wikipedia

Using the approximation, you can rewrite equation $(4)$ as

$$L\frac{d^2\theta}{dt^2} = -g\theta \tag{5}$$

The above equation looks quite similar to the equation $(1)$. It does match perfectly. Therefore, for small amplitudes, the pendulum executes a simple harmonic motion with a reasonable uncertainty.

Does a spring-mass system execute simple harmonic motion?

If the spring obeys Hooke's law, then it always executes simple harmonic motion.

Hooke's law states that:

$$F_{restoring} = ma = - kx \tag{6}$$

It is clearly evident from the above equation that the acceleration is directly proportional to the displacement and acts in the direction opposite to the displacement.

Why do we limit the amplitude of a spring-mass system?

Under high strain, the spring does not obey Hooke's law. This is kinda obvious: if you stretch a spring too much, it deforms permanently. Therefore, the equation $(6)$ no longer holds. If that equation does not hold, then the mass won't execute simple harmonic motion. 

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    $\begingroup$ Springs obey Hooke's law because, and to the extent that, they are carefully designed and manufactured to obey Hooke's law. $\endgroup$
    – Ben
    Commented Feb 21, 2017 at 17:06
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    $\begingroup$ @YashasSamaga, correct. Point: It's not a physical law in the same way as Newton's laws. You can buy nonlinear springs. $\endgroup$
    – Ben
    Commented Feb 21, 2017 at 17:47
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    $\begingroup$ Just for the record, it's not really only under high strain that a spring doesn't obey Hooke's law. A spring never really follows Hooke's law perfectly. There is obviously some non-linear behavior in the spring anyways due to design, material inconsistencies, actual physical behaviour, etc. Springs also just happen to have very linear-elastic behaviour if properly designed and operated in the right range. The spring equation is approximate just like the pendulum equation though (especially since springs work because of shape, it's easy to imagine imperfections giving non-linear $k$). $\endgroup$
    – JMac
    Commented Feb 21, 2017 at 18:48
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    $\begingroup$ @krs013 For $\frac{\pi}{10}$, the approximation is accurate up to the second decimal place. I added a graph to clear things up. $\endgroup$
    – Yashas
    Commented Feb 22, 2017 at 5:08
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    $\begingroup$ @Ben I imagine there are physical laws that underpin the approximation that is Hooke's Law - in the sense that bulk properties of materials of a particular type are linear for "small" deformations - due to intermolecular/interatomic forces. $\endgroup$
    – MikeW
    Commented Feb 24, 2017 at 14:54

Why does a simple pendulum or a spring-mass system show simple harmonic motion (SHM) only for small amplitudes?

Simple harmonic motion (in one dimension) is, by definition, a solution to the generic equation $$\frac{d^2x}{dt^2}+\omega^2x=0,$$ where $x$ is a generic variable (it can be for instance a displacement or an angle). This equation of motion can be obtained from Newton's second law $$m\frac{d^2x}{dt^2}=F=-\frac{dU}{dx},$$ where the last equal sign holds for conservative forces, with $U$ being the potential energy.

Therefore we can compare $$\frac{d^2x}{dt^2}+\frac{1}{m}\frac{dU}{dx}=0,$$ with the equation defining the SHM and see that this requires the potential $U$ to be quadratic in $x$.

Neither a simple pendulum or a real spring has potential which is quadratic (or parabolic) around an equilibrium point. However in a small enough region around the equilibrium (zero force and minimum of the potential) we can Taylor expand the potential up to second order which just gives a parabolic potential. To understand this, consider the figure below,

enter image description here

This is a generic potential which is definitely not of the required form for a SHM. However, notice that we can fit a parabola (dotted line) around any stable equilibrium point such as $x_0$. In the vicinity of the equilibrium point $x_0$ we have the expansion, $$U(x)=U(x_0)+\frac{dU(x_0)}{dx}(x-x_0)+\frac{1}{2}\frac{d^2U(x_0)}{dx^2}(x-x_0)^2+\mathcal O(\Delta x^3),$$ where $\mathcal O(\Delta x^2)$ means we are neglecting terms of order $(x-x_0)^3$ or greater. Since the first derivative of the potential at $x_0$ gives the force at $x_0$ (which is zero), that term vanishes. We are left with $$U(x)=U(x_0)+\frac 12 k(x-x_0)^2++\mathcal O(\Delta x^3),$$ and since $U(x_0)$ is an irrelevant additive term, and $k=frac{d^2U(x_0)}{dx^2}$ is constant, this quadratic potential is exactly the characteristic potential of SHM. Nearby $x_0$ it gives a linear and restoring force. Note however that this approximation does not hold for arbitrary displacements. At some point $x-x_0$ is so large that higher order terms have to be taken into account.

For a simple pendulum, the potential is $-mgl\cos x$ ($x$ being an angle) whose expansion up two order two around the equilibrium $x=0$ gives a parabola. For a real spring, the potential can be even more complicated, in general it is a sum of terms $a_nx^{2n}$. Then for a small region we can drop higher order terms and keep only the leading one $a_1x^2$ which gives the Hook Law.


It just means that the pendulum will only execute SHM with small angles for which the pendulum subtends as it oscillates.

For larger angles the motion of the pendulum deviates from being simple harmonic; that's why the small angle approximation is required in the derivation.


Even for a horizontal mass-spring system the displacement of the mass from its equilibrium position cannot be made too large, otherwise the simple-harmonic motion ceases to be valid.

But whether considering a pendulum undergoing SHM or a mass-spring system the displacement from equilibrium position is $\propto$ to the negative of the acceleration towards the equilibrium point as these are in opposite directions which is the reason for the minus sign.

  • $\begingroup$ When you say "displacement of the mass from its equilibrium position cannot be made too large, otherwise the simple-harmonic motion ceases to be valid" aren't you just assuming a physical spring that has strain limits? Wouldn't an ideal system ignore those? $\endgroup$ Commented Feb 21, 2017 at 13:33
  • $\begingroup$ @Mike I was considering the strain limits on the spring mostly. But if you displace the mass too far the motion will not be 'smooth'. It will be jerky and erratic $\endgroup$
    – user138066
    Commented Feb 21, 2017 at 13:38
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    $\begingroup$ @user395550: Why? Assuming the surface is frictionless. $\endgroup$ Commented Feb 21, 2017 at 18:54

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