Why pendulum does not follow SHM for larger angular displacement?

Question

Considering an ideal case(neglecting drag of air, damping etc.), a pendulum follows SHM if the angular displacement is small (upto 10 degrees). But, for large angular displacement(more than 10 degree), it does not follow SHM. Why? What force cause the pendulum not to follow SHM?

Answer 1

What is SHM?

When the motion of a system is caused by a force that is linearly proportional to displacement from the equilibrium, then it incurs SHM. That is when the body moves in a sinusoidal way, it is in harmonic motion.

Consider a spring whose one end is attached with a bob of mass $m$. Let the bob be compressed & then released. It will oscillate. How to represent it mathematically? It can be represented by a second-order differential equation

$$m\frac{d^2 x}{dt^2} = F(x) = -(k_1x + k_2x^2 + k_3x^3 + k_4x^4 + \cdots).$$

Now, is the motion in SHM? No, because the force is not linearly proportional to $x$. However, we can find a range of values of $x$ for which the sum of the terms not linear in $x$ becomes negligible unless $k_1$ itself is zero.

So, we can write then

$$m\frac{d^2 x}{dt^2} = F(x) = -k_1x.$$ So, this occurs only for certain $x$.

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Pendulum in SHM for small angles:

Let a pendulum having a bob of mass $m$ & massless string of length $l$ is displaced by $\psi(t)$. then from Newton's second law, we get

$$\frac{Ml d^2 \psi}{dt^2} = -Mg\sin\psi(t) \\ \implies \frac{d^2 \psi}{dt^2} = -\frac{g}{l} \left(\psi - \frac{\psi^3}{3!} + \frac{\psi^5}{5!}+ \cdots\right)^1.$$

This is not in SHM as the restoring force is not linearly proportional to $\psi$. So, there must be a range of $\psi$ for which

$$\frac{d^2 \psi}{dt^2} = -\frac{g}{l} \psi;$$

then only can SHM occur & this occurs for small values of $\psi$.

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$^1$ by Taylor series at $x_0 = 0$ or Maclaurin series: $$f(x)= \sum_{n=0}^ {\infty} \frac{f^{(n)} (x_0)}{n!} (x - x_0)^n.$$