In newtonian mechanics equation of motion of a simple pendulum:


And then I approximated for small angles $\sin\theta\simeq\theta$ that yields the equation of simple harmonic motion we all know:


Out of curiosity I decided to derive the equation through Lagrangian mechanics to understand the way the small angle approximation works for the lagrangian:


Then, I realised the small angles approximation for cosine had to be $\cos\theta\simeq 1-\frac{\theta^2}{2}$ instead of $\cos\theta\simeq1$ so I needed the second order approximation to get the simple harmonic motion equation. With some basic calculus I found out that for small angles the error we get approximating cosine to 1 is way bigger than the error we get approximating sine to the first order and they are of the same order if I approximate cosine to second order and sine to first order (that's reasonable since first order sine expansion is the same as second order expansion). My question is: Why when we derive the equation of a wave on a string (assuming constant tension and small angles, elastic waves, and constant linear density), we neglect the horizontal force acting on an element of string? I'll write down Newton's second law for a piece of string of mass $\Delta m$: Let $\tau$ be the tension of the rope.

$$\vec{F}=\Delta m \vec{a}$$


Tensions acting at its ends will have the same magnitudes, so we get:

$$\tau(\cos\theta_2-\cos\theta_1)=\Delta m a_x$$

$$\tau(\sin\theta_2-\sin\theta_1)=\Delta m a_y$$

Without going further with this derivation of D'Alembert equation, I've read some books approximating $\sin\theta\simeq\theta$ and $\cos\theta\simeq1$ (so $a_x\simeq0$). If we were to expand cosine to second order (like I said before), would we also get longitudinal waves? If not, why does this approximation work for this model and not for simple pendulum?

  • $\begingroup$ My guess is that we neglect the horizontal force because the horizontal motion of the string element is much smaller than the vertical. But when examining 3D resonators (eg chime bars & bells) we do need to take the longitudinal waves into account. $\endgroup$
    – PM 2Ring
    Apr 23, 2020 at 10:31

1 Answer 1



The reason why the approximation doesn't work in the pendulum case is because you are applying it at the wrong place.

Correct way

You should apply the approximation after you differentiate the Lagrangian when applying the Euler-Lagrange equations. Thus

\begin{align} \frac{\mathrm d}{\mathrm d t} \left(\frac{\mathrm d \mathcal L}{\mathrm d \dot{\theta}}\right)&=\frac{\mathrm d \mathcal L }{\mathrm d \theta}\\[5pt] ml^2 \ddot{\theta}&=-mgl \sin\theta \end{align}

Now you can apply the approximation that $\sin\theta \approx \theta$, thus


which is what you expected.

Fallacy in your argument

The reason why we need to include the second order ($-\theta^2/2$) while approximating $\cos \theta$ is because we are going to differentiate that expression. And once we differentiate the expression, the second order term becomes a first order term ($-\theta$) and thus it suddenly becomes "important". Excluding it, would give us a useless and wrong solution. But however in the string wave case, we aren't going to use any operation which might turn the second order term into a significant first or zero order term. Thus it makes sense not to include that second order term in the derivation.


Always take all the approximations once you've finished applying all the operations which might involve a change in the order (exponent/powers) of the terms. In fact, you should always use the complete Taylor expansion of any function until you get your final expression. This idea is really important and needs to be kept in mind when treating small quantities (as in your case, $\theta$).

  • $\begingroup$ I see now. So, when I want to approximate the lagrangian before differentiating I need to take into account second order terms that will become first order terms. Once we have the equation of motion, the second order terms must be neglected. I guess my question was a stupid one. Thanks for clarifying. $\endgroup$ Apr 23, 2020 at 12:21
  • 1
    $\begingroup$ @Feynman_00 Yes, exactly and the question was not at all stupid. BTW, if you weren't aware, then you can accept the answer. $\endgroup$
    – user258881
    Apr 23, 2020 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.