# Derivation of time period of simple pendulum given by $T=2 \pi \sqrt{\frac{l}{g}}$

I was doing a few problems on the time period of time period of simple pendulum by applying $$T=2 \pi \sqrt{\dfrac{l}{g}}$$, but I needed to know the derivation also not given in my textbook. Please help me with the derivation of the time period of a simple pendulum. I think it looks similar to the time period of SHM i.e. $$T=2 \pi \sqrt{\dfrac{m}{k}}$$, which I studied in past classes.

Let,
$$l=$$ length of pendulum
$$x=$$ amount of pendulum bob displaced
$$g=$$ acceleration due to gravity
$$θ=$$ angle subtended by the pendulum with the vertical

So as we know that the general equation for SHM

$$A=-\omega^2x\tag1$$

So, here we have to break the force vector in its components.

So, from the diagram we get to know that the horizontal component of gravitational force is getting cancelled with the tension as the tension force is self adjusting in nature.

Hence we get driving force and that driving force is the vertical component of the gravirational force.

$$F_{\text{net}}=Mg\sin\theta$$

$$Ma=Mg\sin\theta$$

Cancelling $$M$$ on both sides as Mass can't be zero

$$a=g\sin\theta$$

From the diagram,

$$\sin\theta=\frac xl$$

By substituting this we get,

$$A=g\frac xl$$

Putting a -ve sign because as we displace the bob let's say in the right the acceleration acts exactly in its opposite direction so,

$$A=-g\frac xl$$

Substituting from equation $$(1)$$ we get,

$$-\omega^2x=-g\frac xl$$

Cancelling -ve sign and $$x$$ from both the sides we get,

$$\omega=\sqrt{\frac gl}$$

So as we know that general time period formula is

$$T=\frac{2π}{\omega}$$

By substituting $$\omega$$ we get,

$$T=\frac{2π}{\sqrt{g/l}}$$

Final equation will be,

$$T=2π \sqrt{\frac lg}$$

Yay we proved it!!

• What does SHM stand for? Commented Jan 10, 2022 at 8:00
• It's Simple Harmonic Motion. Nothing but just a special case of Harmonic motion. Commented Jan 10, 2022 at 8:02
• Why is T = 2π/ω instead of T = 2θ/ω?
– Shub
Commented Jan 20, 2022 at 16:25
• $T$ is meant for time period of a periodic motion it is the specific time after which the motion of the body starts repeating the pattern itself. So let's take a simple example of circular motion and assuming some point at which the motion of the body was initially started with some constant angular velocity $ω$ and after this hence it's a circular motion so it is going to repeat this motion again and again so what would be the angular displacement after which the body will repeat it's motion again and again? So that's $2π$ or 360⁰ so how much time will it require to make this amount of..... Commented Jan 20, 2022 at 16:36
• Angular displacement that is by using the equation of the angular velocity that's $ω = \frac{θ}{t}$ and in this case $θ$ is 2π so putting this in this equation you will get $t = \frac{2π}{ω}$ So because it's a special thing so we give it a notation $T$ instead of using $t$. So here you got this thing! Commented Jan 20, 2022 at 16:42

The equation of motion describing a pendulum is given by:

$$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin(\theta)$$

For small angles, one can approximate:

$$\frac{d^2\theta}{dt^2} \approx -\frac{g}{L}\theta$$

Solving this equation, one finds the general solution to be:

$$\theta(t) = A\sin\left(\sqrt{\frac gL}t\right) + B\cos\left(\sqrt{\frac gL}t\right)$$

With $$A$$ and $$B$$ some constants depending on the initial condition. If we choose $$\theta(t = 0) = 0$$ (which is still general, since the pendulum swings the whole time, we just choose here were we start to measure it you could say), we find that $$B = 0$$, so:

$$\theta(t) = A\sin\left(\sqrt{\frac gL}t\right)$$

Now what is the period of $$A\sin(\sqrt{g/L} \ t)$$? Indeed, it is:

$$T = 2\pi\sqrt{\frac Lg}$$

• the way you set this up it should be $B=0$ Commented Jan 9, 2022 at 8:15
• For a fixed initial condition, how do we prove that the solution of $$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\theta$$ is a good approximation of the solution of $$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin(\theta)$$? Commented Jan 9, 2022 at 11:23
• Hi Filippo, it is only a good approximation at small angles, but this has nothing to do with the initial condition I choose. The system is oscillating back on forth all the time, so you can always do a time translation which reflects another initial condition. It is however important what the value of B is, since the maximal angle is fount to be B (when the cosine is 1), so B should be small enough for this approximation to be valid. Commented Jan 9, 2022 at 11:32
• "It is only a good approximation at small angles, but this has nothing to do with the initial condition I choose" - Intuitively, if the initial angle/velocity is "too big", then the pendulum will reach angles where $\sin\theta=\theta$ is not a good approximation. So I do believe that the initial condition does matter. However, I See that this is precisely what you say in your last sentence, so maybe I have misunderstood your comment. Commented Jan 9, 2022 at 12:55
• "The system is oscillating back on forth all the time, so you can always do a time translation which reflects another initial condition" - How do you know that the solution of $$\frac{d^2\theta}{dt^2} = -\frac{g}{L}\sin(\theta)$$is periodic? Commented Jan 9, 2022 at 13:41